Advanced Math

438 soru

Soru 281Soru

What is the larger solution to the equation below?

xx2+3x=2\frac{x}{x - 2} + \frac{3}{x} = 2
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Cevap: 6

Cevap

The larger solution to the equation is 6.
To solve the rational equation, we clear the denominators by multiplying both sides by x(x2)x(x - 2), which yields the quadratic equation x27x+6=0x^2 - 7x + 6 = 0. Factoring this equation gives (x6)(x1)=0(x - 6)(x - 1) = 0, leading to the solutions x=6x = 6 and x=1x = 1. Both values are valid because they do not make any denominator of the original expression equal to zero. The larger of the two values is 6.

Adım Adım Çözüm

1
Multiply both sides of the equation by the common denominator x(x2)x(x - 2) to clear the fractions.
x2+3(x2)=2x(x2)x^2 + 3(x - 2) = 2x(x - 2)
Clearing denominators simplifies the rational equation into a polynomial equation.
2
Expand both sides of the equation.
x2+3x6=2x24xx^2 + 3x - 6 = 2x^2 - 4x
Distributing the multiplication allows us to combine like terms.
3
Move all terms to one side of the equation to write it in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x27x+6=0x^2 - 7x + 6 = 0
Setting the quadratic expression to zero prepares it for factoring.
4
Factor the quadratic equation.
(x6)(x1)=0(x - 6)(x - 1) = 0
Finding two numbers that multiply to 66 and sum to 7-7 gives 6-6 and 1-1.
5
Identify the values of xx that satisfy the factored equation and check for extraneous solutions.
x=6x = 6 or x=1x = 1
Neither solution makes the denominators in the original equation, x2x-2 or xx, equal to zero, so both are valid. The larger of these two solutions is 6.

Anahtar Kavram

Solving rational equations by clearing denominators to form a quadratic equation, and verifying solutions against the domain constraints.
Soru 282Soru
For all real values of xx that satisfy the equation below, what is the value of x+1x + 1?
5x+393=x\sqrt{5x + 39} - 3 = x
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Cevap: 66

Cevap

The correct answer is 66.
By adding 33 to both sides, we get 5x+39=x+3\sqrt{5x + 39} = x + 3. Squaring both sides yields 5x+39=x2+6x+95x + 39 = x^2 + 6x + 9. Setting this quadratic equation to zero gives x2+x30=0x^2 + x - 30 = 0, which factors as (x+6)(x5)=0(x + 6)(x - 5) = 0. The potential solutions are x=5x = 5 and x=6x = -6. Checking these values in the original equation shows that only x=5x = 5 is valid: 5(5)+393=83=5\sqrt{5(5) + 39} - 3 = 8 - 3 = 5, which satisfies the equation. The value x=6x = -6 is extraneous since 5(6)+393=33=06\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6. Therefore, the value of x+1x + 1 is 5+1=65 + 1 = 6.

Adım Adım Çözüm

1
Isolate the radical expression on one side of the equation.
5x+39=x+3\sqrt{5x + 39} = x + 3
Isolating the radical allows us to square both sides to eliminate the square root.
2
Square both sides of the equation.
5x+39=(x+3)2    5x+39=x2+6x+95x + 39 = (x + 3)^2 \implies 5x + 39 = x^2 + 6x + 9
Squaring a square root removes the radical, allowing us to solve the resulting quadratic equation.
3
Rearrange the terms to form a standard quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0.
x2+x30=0x^2 + x - 30 = 0
Subtracting 5x5x and 3939 from both sides sets the quadratic expression equal to zero.
4
Factor the quadratic equation to find potential solutions.
(x+6)(x5)=0    x=6 or x=5(x + 6)(x - 5) = 0 \implies x = -6 \text{ or } x = 5
Finding two numbers that multiply to 30-30 and add up to 11 gives 66 and 5-5.
5
Substitute both candidate solutions back into the original equation to check for extraneous solutions.
For x=5x = 5: 5(5)+393=83=5\sqrt{5(5) + 39} - 3 = 8 - 3 = 5 (Valid). For x=6x = -6: 5(6)+393=33=06\sqrt{5(-6) + 39} - 3 = 3 - 3 = 0 \neq -6 (Extraneous).
Squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original relation.
6
Use the valid solution to calculate the requested expression.
x+1=5+1=6x + 1 = 5 + 1 = 6
The question asks for the value of x+1x + 1.

Anahtar Kavram

Solving radical equations and identifying extraneous solutions
Soru 283Soru

If 16x+12x3=64\frac{16^{x+1}}{2^{x-3}} = 64, what is the value of xx?

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Cevap: 13-\frac{1}{3}

Cevap

13-\frac{1}{3}
The correct answer is 13-\frac{1}{3}. By expressing all terms with base 2, the equation 16x+12x3=64\frac{16^{x+1}}{2^{x-3}} = 64 becomes 24x+42x3=26\frac{2^{4x+4}}{2^{x-3}} = 2^6. Applying exponent division rules yields 2(4x+4)(x3)=262^{(4x+4)-(x-3)} = 2^6, which simplifies to 23x+7=262^{3x+7} = 2^6. Setting the exponents equal gives 3x+7=63x + 7 = 6, which simplifies to x=13x = -\frac{1}{3}.

Adım Adım Çözüm

1
Rewrite all terms in the equation using a common base of 2.
The equation becomes (24)x+12x3=26\frac{(2^4)^{x+1}}{2^{x-3}} = 2^6.
Expressing bases 16 and 64 as powers of 2 allows us to apply exponent laws to simplify the equation.
2
Apply the power of a power rule to the numerator, then apply the quotient rule for exponents.
The left side simplifies to 24x+4(x3)=23x+72^{4x+4 - (x-3)} = 2^{3x+7}.
The expression (24)x+1(2^4)^{x+1} becomes 24x+42^{4x+4}. Using the quotient rule aman=amn\frac{a^m}{a^n} = a^{m-n}, we subtract the denominator's exponent (x3)(x-3) from the numerator's exponent (4x+4)(4x+4), distributing the negative sign to get 3x+73x+7.
3
Set the exponents equal to each other and solve the resulting linear equation.
3x+7=6    3x=1    x=133x + 7 = 6 \implies 3x = -1 \implies x = -\frac{1}{3}.
Since the bases on both sides of the equation are equal, their exponents must be equal.

Anahtar Kavram

Solving exponential equations by rewriting with a common base
Tahmini Süre:1m 30s
Soru 284Soru

The quadratic function ff is defined by f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are constants. In the xyxy-plane, the graph of ff is a parabola with vertex (3,5)(3, -5) that passes through the point (0,4)(0, 4). What is the value of a+b+ca + b + c?

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Cevap: -1

Cevap

-1
The vertex form of a quadratic function is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex. Substituting the given vertex (3,5)(3, -5) gives f(x)=a(x3)25f(x) = a(x - 3)^2 - 5. Since the graph passes through the point (0,4)(0, 4), substituting x=0x = 0 and f(x)=4f(x) = 4 yields the equation 4=a(03)254 = a(0 - 3)^2 - 5, which simplifies to 9a=99a = 9, or a=1a = 1. The function is therefore defined by f(x)=(x3)25f(x) = (x - 3)^2 - 5. The expression a+b+ca + b + c represents the sum of the coefficients of the quadratic function in standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c. Evaluating the function at x=1x = 1 gives f(1)=a(1)2+b(1)+c=a+b+cf(1) = a(1)^2 + b(1) + c = a + b + c. Substituting x=1x = 1 into our vertex form equation yields f(1)=(13)25=(2)25=45=1f(1) = (1 - 3)^2 - 5 = (-2)^2 - 5 = 4 - 5 = -1. Therefore, the value of a+b+ca + b + c is 1-1.

Adım Adım Çözüm

1
Write the quadratic function in vertex form using the given vertex (3,5)(3, -5).
f(x)=a(x3)25f(x) = a(x - 3)^2 - 5, where aa is a constant.
The vertex form of a quadratic function is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
2
Substitute the coordinates of the given point (0,4)(0, 4) into the vertex form equation to solve for aa.
4=a(03)254 = a(0 - 3)^2 - 5, which simplifies to 4=9a54 = 9a - 5, and solving for aa gives a=1a = 1.
Since the graph passes through (0,4)(0, 4), substituting these coordinates into the function's equation must yield a true statement.
3
Find the value of a+b+ca + b + c by evaluating f(1)f(1).
f(1)=1(13)25=1(2)25=45=1f(1) = 1(1 - 3)^2 - 5 = 1(-2)^2 - 5 = 4 - 5 = -1.
For any quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c, evaluating the function at x=1x = 1 gives f(1)=a(1)2+b(1)+c=a+b+cf(1) = a(1)^2 + b(1) + c = a + b + c.

Anahtar Kavram

Quadratic Functions and Graphs
Soru 285Soru

In the xyxy-plane, the graph of a cubic polynomial function pp with real coefficients has exactly two xx-intercepts, at (1,0)(1, 0) and (4,0)(4, 0). If the graph of pp passes through the points (0,8)(0, -8) and (2,2)(2, 2), what is the value of p(6)p(6)?

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Cevap: 10

Cevap

10
A cubic polynomial with real coefficients and exactly two xx-intercepts at (1,0)(1, 0) and (4,0)(4, 0) must have one root of multiplicity 1 and one root of multiplicity 2. This yields two possible forms: p(x)=a(x1)2(x4)p(x) = a(x - 1)^2(x - 4) or p(x)=a(x1)(x4)2p(x) = a(x - 1)(x - 4)^2. Substituting the yy-intercept (0,8)(0, -8) into the first form gives a=2a = 2, but the resulting polynomial p(x)=2(x1)2(x4)p(x) = 2(x - 1)^2(x - 4) does not pass through (2,2)(2, 2) since p(2)=4p(2) = -4. Substituting (0,8)(0, -8) into the second form gives a=12a = \frac{1}{2}, and the resulting polynomial p(x)=12(x1)(x4)2p(x) = \frac{1}{2}(x - 1)(x - 4)^2 correctly passes through (2,2)(2, 2) since p(2)=2p(2) = 2. Finally, evaluating this function at x=6x = 6 yields p(6)=12(61)(64)2=10p(6) = \frac{1}{2}(6 - 1)(6 - 4)^2 = 10.

Adım Adım Çözüm

1
Determine the possible forms of the cubic polynomial based on its xx-intercepts.
Two possible cases: Case 1: p(x)=a(x1)2(x4)p(x) = a(x - 1)^2(x - 4) or Case 2: p(x)=a(x1)(x4)2p(x) = a(x - 1)(x - 4)^2, where aa is a non-zero constant.
Since the polynomial has degree 3 and exactly two distinct xx-intercepts, one intercept must be a root of multiplicity 2 (tangent to the xx-axis) and the other must be a root of multiplicity 1 (crossing the xx-axis).
2
Solve for the constant aa in Case 1 using the yy-intercept (0,8)(0, -8).
a=2a = 2, yielding the candidate function p(x)=2(x1)2(x4)p(x) = 2(x - 1)^2(x - 4).
Substituting x=0x = 0 and p(0)=8p(0) = -8 into the equation for Case 1 allows us to solve for aa.
3
Test the point (2,2)(2, 2) in the Case 1 candidate function.
p(2)=42p(2) = -4 \neq 2, which means Case 1 is incorrect.
The correct function must satisfy all given points, including (2,2)(2, 2).
4
Solve for the constant aa in Case 2 using the yy-intercept (0,8)(0, -8).
a=12a = \frac{1}{2}, yielding the candidate function p(x)=12(x1)(x4)2p(x) = \frac{1}{2}(x - 1)(x - 4)^2.
Substituting x=0x = 0 and p(0)=8p(0) = -8 into the equation for Case 2 allows us to solve for aa.
5
Test the point (2,2)(2, 2) in the Case 2 candidate function.
p(2)=2p(2) = 2, which matches the given point.
Confirming that the Case 2 function is consistent with the point (2,2)(2, 2).
6
Evaluate the correct function at x=6x = 6.
p(6)=10p(6) = 10.
Substituting x=6x = 6 into the verified function p(x)=12(x1)(x4)2p(x) = \frac{1}{2}(x - 1)(x - 4)^2 to find the final answer.

Anahtar Kavram

Identifying the algebraic form of a polynomial from its xx-intercepts and multiplicities, and determining unknown coefficients using coordinate points.
Soru 286Soru

If the graph of y=f(x)y = f(x) contains the point (3,7)(3, 7), and the function gg is defined by g(x)=f(x+4)2g(x) = f(x + 4) - 2, what is the value of g(1)g(-1)?

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Cevap: 5

Cevap

5
Since the graph of y=f(x)y = f(x) contains the point (3,7)(3, 7), we have f(3)=7f(3) = 7. The function gg is defined as g(x)=f(x+4)2g(x) = f(x + 4) - 2. To find the value of g(1)g(-1), we substitute x=1x = -1 into the definition of gg: g(1)=f(1+4)2=f(3)2g(-1) = f(-1 + 4) - 2 = f(3) - 2. Substituting f(3)=7f(3) = 7 gives g(1)=72=5g(-1) = 7 - 2 = 5.

Adım Adım Çözüm

1
Translate the point (3,7)(3, 7) on the graph of f(x)f(x) into function notation.
f(3)=7f(3) = 7
By definition, if a point (a,b)(a, b) is on the graph of y=f(x)y = f(x), then f(a)=bf(a) = b.
2
Substitute x=1x = -1 into the expression for g(x)g(x) to evaluate g(1)g(-1).
g(1)=f(1+4)2g(-1) = f(-1 + 4) - 2
To find g(1)g(-1), replace every occurrence of xx with 1-1 in the function definition of g(x)g(x).
3
Simplify the input of the function ff and compute the final value.
g(1)=f(3)2=72=5g(-1) = f(3) - 2 = 7 - 2 = 5
Simplify 1+4-1 + 4 to 33, then substitute the known value f(3)=7f(3) = 7 and subtract 22.

Anahtar Kavram

Evaluating a transformed function using function notation and given coordinate points.
Soru 287Soru

In the quadratic equation x28x+k=0x^2 - 8x + k = 0, kk is a constant. If the difference between the two real solutions to the equation is 22, what is the value of kk?

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Cevap: 15

Cevap

15
The correct answer is 15. The sum of the solutions to the quadratic equation x28x+k=0x^2 - 8x + k = 0 is given by 81=8-\frac{-8}{1} = 8. We are given that the difference between the two solutions is 22. Let the two solutions be x1x_1 and x2x_2. We can set up the system of equations: x1+x2=8x_1 + x_2 = 8 and x1x2=2x_1 - x_2 = 2. Adding the equations gives 2x1=102x_1 = 10, so x1=5x_1 = 5. Thus, x2=3x_2 = 3. The product of the solutions is equal to the constant term kk. Therefore, k=x1×x2=5×3=15k = x_1 \times x_2 = 5 \times 3 = 15. Alternatively, using the difference of roots formula, b24aca=2(8)24(1)(k)1=2644k=2\frac{\sqrt{b^2 - 4ac}}{a} = 2 \Rightarrow \frac{\sqrt{(-8)^2 - 4(1)(k)}}{1} = 2 \Rightarrow \sqrt{64 - 4k} = 2. Squaring both sides gives 644k=44k=60k=1564 - 4k = 4 \Rightarrow 4k = 60 \Rightarrow k = 15.

Adım Adım Çözüm

1
Find the sum of the two solutions of the quadratic equation.
The sum of the solutions is 88.
For a quadratic equation in the form x2+bx+c=0x^2 + bx + c = 0, the sum of the solutions is equal to b-b. Here, (8)=8-(-8) = 8.
2
Set up a system of linear equations using the sum and the given difference of the solutions to find the individual solutions.
The solutions are 55 and 33.
Let the solutions be x1x_1 and x2x_2. We have the system: x1+x2=8x_1 + x_2 = 8 and x1x2=2x_1 - x_2 = 2. Adding these two equations gives 2x1=102x_1 = 10, so x1=5x_1 = 5. Substituting 55 back into the first equation gives 5+x2=85 + x_2 = 8, so x2=3x_2 = 3.
3
Use the product of the solutions to determine the value of kk.
The value of kk is 1515.
For a quadratic equation in the form x2+bx+c=0x^2 + bx + c = 0, the product of the solutions is equal to cc, which is kk. Therefore, k=x1×x2=5×3=15k = x_1 \times x_2 = 5 \times 3 = 15.

Anahtar Kavram

Relationship between the roots and coefficients of a quadratic equation (Vieta's Formulas).
Soru 288Soru

A polynomial function pp of degree 3 has xx-intercepts at (2,0)(-2, 0) with multiplicity 2, and (3,0)(3, 0) with multiplicity 1. In the xyxy-plane, the graph of y=p(x)y = p(x) intersects the yy-axis at (0,24)(0, 24). What is the remainder when p(x)p(x) is divided by x1x - 1?

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Cevap: 3636

Cevap

The remainder when the polynomial function is divided by x1x - 1 is 36.
The correct answer is 36. A polynomial with a root at x=2x = -2 of multiplicity 2 and a root at x=3x = 3 of multiplicity 1 has the form p(x)=a(x+2)2(x3)p(x) = a(x + 2)^2(x - 3). Since the yy-intercept is (0,24)(0, 24), we solve p(0)=a(2)2(3)=24p(0) = a(2)^2(-3) = 24 to find a=2a = -2. Thus, the polynomial is p(x)=2(x+2)2(x3)p(x) = -2(x + 2)^2(x - 3). According to the Remainder Theorem, dividing p(x)p(x) by x1x - 1 leaves a remainder of p(1)p(1). Substituting x=1x = 1 yields p(1)=2(3)2(2)=36p(1) = -2(3)^2(-2) = 36.

Adım Adım Çözüm

1
Write the general form of the cubic polynomial using its roots and multiplicities.
p(x)=a(x+2)2(x3)p(x) = a(x + 2)^2(x - 3)
Since there is an xx-intercept at x=2x = -2 with multiplicity 2, (x+2)2(x + 2)^2 is a factor. Since there is an xx-intercept at x=3x = 3 with multiplicity 1, (x3)(x - 3) is a factor. Here, aa is a constant coefficient.
2
Determine the value of the constant coefficient aa using the yy-intercept.
a=2a = -2, so p(x)=2(x+2)2(x3)p(x) = -2(x + 2)^2(x - 3)
The graph intersects the yy-axis at (0,24)(0, 24), meaning p(0)=24p(0) = 24. Substituting x=0x = 0 gives p(0)=a(0+2)2(03)=12ap(0) = a(0 + 2)^2(0 - 3) = -12a. Setting 12a=24-12a = 24 yields a=2a = -2.
3
Apply the Remainder Theorem to find the required remainder.
The remainder is equal to p(1)p(1).
By the Remainder Theorem, the remainder when a polynomial p(x)p(x) is divided by xcx - c is p(c)p(c). Here, the divisor is x1x - 1, so we evaluate the polynomial at x=1x = 1.
4
Calculate the value of p(1)p(1).
p(1)=36p(1) = 36
Substituting x=1x = 1 into p(x)=2(x+2)2(x3)p(x) = -2(x + 2)^2(x - 3) gives p(1)=2(1+2)2(13)=2(9)(2)=36p(1) = -2(1 + 2)^2(1 - 3) = -2(9)(-2) = 36.

Anahtar Kavram

Identifying a polynomial from its roots and multiplicities, solving for its leading coefficient using a given point, and applying the Remainder Theorem.
Soru 289Soru

The function ff is defined for all real numbers, and the graph of y=f(x)y = f(x) in the xyxy-plane has a single minimum at the point (5,2)(5, -2). The function gg is defined by g(x)=3f(2x4)+7g(x) = -3f(2x - 4) + 7. What is the yy-coordinate of the maximum point on the graph of y=g(x)y = g(x)?

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Cevap: 13

Cevap

The correct answer is 13.
The graph of y=f(x)y = f(x) has a minimum at (5,2)(5, -2), which means f(5)=2f(5) = -2 and f(x)2f(x) \ge -2 for all xx. The function g(x)=3f(2x4)+7g(x) = -3f(2x-4) + 7 includes a vertical stretch by a factor of 33, a vertical reflection across the xx-axis, and a vertical shift upward by 77 units. Because of the vertical reflection, the minimum value of the original function becomes the maximum value of the transformed function. Applying the vertical transformations to the yy-coordinate of the minimum point yields 3(2)+7=6+7=13-3(-2) + 7 = 6 + 7 = 13.

Adım Adım Çözüm

1
Identify the minimum point and minimum value of the original function f(x)f(x).
f(5)=2f(5) = -2, and f(x)2f(x) \ge -2 for all real numbers xx.
The problem states that the graph of y=f(x)y = f(x) has a single minimum at the point (5,2)(5, -2).
2
Determine the transformed xx-coordinate corresponding to the original input of 55.
2x4=5    2x=9    x=4.52x - 4 = 5 \implies 2x = 9 \implies x = 4.5.
Setting the argument of the function f(2x4)f(2x-4) equal to the original minimum input of 55 allows us to find the corresponding input xx for the function gg.
3
Apply the vertical transformations to find the output value of g(x)g(x) at x=4.5x = 4.5.
g(4.5)=3f(5)+7=3(2)+7=6+7=13g(4.5) = -3f(5) + 7 = -3(-2) + 7 = 6 + 7 = 13.
Substituting f(5)=2f(5) = -2 into the definition of g(x)g(x) gives the vertical transformation of the point.
4
Confirm that the point is indeed the maximum of the transformed function g(x)g(x).
Since f(2x4)2f(2x-4) \ge -2, multiplying by 3-3 yields 3f(2x4)6-3f(2x-4) \le 6. Adding 77 yields g(x)13g(x) \le 13, confirming that 1313 is the maximum value.
Multiplying a function by a negative number reflects its graph vertically, changing a minimum point into a maximum point.

Anahtar Kavram

Applying horizontal and vertical transformations to function coordinates, and understanding how vertical reflections affect the extrema (minima and maxima) of a graph.
Soru 290Soru

In the xyxy-plane, the graph of the quadratic function f(x)=x2+bx+cf(x) = -x^2 + bx + c, where bb and cc are constants, has its vertex at (h,k)(h, k). The function gg is defined by g(x)=f(x3)+4g(x) = f(x - 3) + 4. The graph of gg passes through the origin (0,0)(0,0), and its vertex lies on the line y=xy = x in the first quadrant. What is the value of f(0)f(0)?

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Cevap: -7

Cevap

-7
The correct answer is 7-7. Representing f(x)f(x) in vertex form as f(x)=(xh)2+kf(x) = -(x-h)^2 + k tells us that translating the function 3 units right and 4 units up shifts the vertex from (h,k)(h, k) to (h+3,k+4)(h+3, k+4). Since this vertex lies on the line y=xy=x, we have h+3=k+4h+3 = k+4, or k=h1k = h-1. Since the vertex of gg lies in the first quadrant, its coordinates must be positive, meaning h>3h > -3 and k>4k > -4. Using the fact that the graph of gg passes through the origin, we have g(0)=0    f(3)+4=0    f(3)=4g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4. Substituting x=3x = -3 into the vertex form of f(x)f(x) gives (3h)2+k=4    (h+3)2k=4-(-3-h)^2 + k = -4 \implies (h+3)^2 - k = 4. Substituting k=h1k = h-1 yields (h+3)2(h1)=4    h2+5h+6=0(h+3)^2 - (h-1) = 4 \implies h^2 + 5h + 6 = 0, which factors into (h+2)(h+3)=0(h+2)(h+3) = 0. The solution h=3h = -3 is discarded because it places the vertex of gg at (0,0)(0,0), which is not in the first quadrant. Therefore, h=2h = -2 and k=3k = -3. Calculating f(0)f(0) gives h2+k=(2)2+(3)=7-h^2 + k = -(-2)^2 + (-3) = -7.

Adım Adım Çözüm

1
Determine the vertex form of f(x)f(x) and the vertex of g(x)g(x).
Since f(x)=x2+bx+cf(x) = -x^2 + bx + c has its vertex at (h,k)(h, k), its vertex form is f(x)=(xh)2+kf(x) = -(x - h)^2 + k. The function g(x)=f(x3)+4g(x) = f(x - 3) + 4 represents a horizontal shift of ff by 3 units to the right and a vertical shift by 4 units up. Thus, the vertex of gg is (h+3,k+4)(h + 3, k + 4).
Understanding translations allows us to write the coordinates of the new vertex in terms of the original vertex variables.
2
Use the line y=xy = x and quadrant constraints to find a relation between hh and kk.
Since the vertex of gg lies on the line y=xy = x, we set its coordinates equal: h+3=k+4h + 3 = k + 4, which simplifies to k=h1k = h - 1. Additionally, because the vertex is in the first quadrant, we must have h+3>0h + 3 > 0 and k+4>0k + 4 > 0.
The geometric placement of the vertex on the line y=xy = x constrains its coordinate values.
3
Apply the condition that the graph of gg passes through the origin.
g(0)=0    f(3)+4=0    f(3)=4g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4. Substituting x=3x = -3 into the vertex form of f(x)f(x) gives (3h)2+k=4-(-3 - h)^2 + k = -4, which simplifies to (h+3)2k=4(h + 3)^2 - k = 4.
The point (0,0)(0,0) lying on the graph of gg provides an equation to solve for the vertex parameters.
4
Substitute k=h1k = h - 1 into the quadratic equation and solve for hh.
(h+3)2(h1)=4    h2+6h+9h+1=4    h2+5h+6=0(h + 3)^2 - (h - 1) = 4 \implies h^2 + 6h + 9 - h + 1 = 4 \implies h^2 + 5h + 6 = 0. Factoring gives (h+2)(h+3)=0(h + 2)(h + 3) = 0, so h=2h = -2 or h=3h = -3.
Substituting the linear relation into the quadratic equation isolates the variable hh.
5
Filter the solutions using the first quadrant constraint and calculate f(0)f(0).
If h=3h = -3, then k=4k = -4, giving the vertex of gg at (0,0)(0, 0), which is not in the first quadrant. If h=2h = -2, then k=3k = -3, giving the vertex of gg at (1,1)(1, 1), which is in the first quadrant. Thus, h=2h = -2 and k=3k = -3. The value of f(0)f(0) is f(0)=(0h)2+k=h2+k=(2)2+(3)=43=7f(0) = -(0 - h)^2 + k = -h^2 + k = -(-2)^2 + (-3) = -4 - 3 = -7.
The quadrant condition uniquely determines the correct vertex coordinates, allowing us to find the y-intercept of the original function.

Anahtar Kavram

Quadratic Functions and Graphs
Soru 291Soru

If xx satisfies the equation below, what is the value of x3x - 3?

x2x+7=4x - \sqrt{2x + 7} = 4
Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6
The correct answer is the value 66. Isolating the radical in x2x+7=4x - \sqrt{2x + 7} = 4 gives x4=2x+7x - 4 = \sqrt{2x + 7}. Squaring both sides yields x28x+16=2x+7x^2 - 8x + 16 = 2x + 7, which simplifies to the quadratic equation x210x+9=0x^2 - 10x + 9 = 0. Factoring this equation gives (x9)(x1)=0(x - 9)(x - 1) = 0, yielding potential solutions of x=9x = 9 and x=1x = 1. Substituting these potential solutions back into the original equation reveals that x=9x = 9 is a valid solution because 925=49 - \sqrt{25} = 4, whereas x=1x = 1 is extraneous because 19=241 - \sqrt{9} = -2 \neq 4. The value of the expression x3x - 3 is 93=69 - 3 = 6.

Adım Adım Çözüm

1
Isolate the radical term in the equation.
x4=2x+7x - 4 = \sqrt{2x + 7}
Before squaring both sides of a radical equation, it is best to isolate the radical term on one side.
2
Square both sides of the equation to eliminate the radical.
(x4)2=2x+7(x - 4)^2 = 2x + 7
Squaring a square root removes the radical, allowing us to solve the resulting algebraic equation.
3
Expand the squared binomial and move all terms to one side to set the quadratic equation to zero.
x210x+9=0x^2 - 10x + 9 = 0
(x4)2(x - 4)^2 expands to x28x+16x^2 - 8x + 16. Subtracting 2x2x and 77 from both sides gives the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Factor the quadratic equation.
(x9)(x1)=0(x - 9)(x - 1) = 0
Finding two numbers that multiply to 99 and add to 10-10 gives 9-9 and 1-1. Thus, the potential solutions are x=9x = 9 and x=1x = 1.
5
Substitute both potential solutions back into the original equation to check for extraneous solutions.
x=9x = 9 is valid, while x=1x = 1 is extraneous.
Checking x=9x = 9: 92(9)+7=95=49 - \sqrt{2(9)+7} = 9 - 5 = 4 (true). Checking x=1x = 1: 12(1)+7=13=241 - \sqrt{2(1)+7} = 1 - 3 = -2 \neq 4 (false). Thus, x=9x = 9 is the only real solution.
6
Calculate the value of the requested expression x3x - 3.
93=69 - 3 = 6
Since the only valid solution is x=9x = 9, we substitute 99 into the expression x3x - 3 to find the final answer.

Anahtar Kavram

Radical equations require isolating the radical, squaring both sides, solving the resulting equation, and checking for extraneous solutions that do not satisfy the original equation.
Tahmini Süre:1m 30s
Soru 292Soru

In the xyxy-plane, the graph of the quadratic function f(x)=x24x5f(x) = x^2 - 4x - 5 intersects the xx-axis at the points (p,0)(p, 0) and (q,0)(q, 0) and has vertex (h,k)(h, k). What is the area of the triangle with vertices at (p,0)(p, 0), (q,0)(q, 0), and (h,k)(h, k)?

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Cevap: 27

Cevap

The area of the triangle is 27.
To find the area of the triangle, we first determine the coordinates of its vertices. The base of the triangle lies on the xx-axis, with endpoints at the xx-intercepts of the function f(x)=x24x5f(x) = x^2 - 4x - 5. Solving x24x5=0x^2 - 4x - 5 = 0 by factoring gives (x5)(x+1)=0(x - 5)(x + 1) = 0, so the intercepts are at x=1x = -1 and x=5x = 5. The distance between these two points is 5(1)=65 - (-1) = 6, which is the base of the triangle. The third vertex is the vertex of the parabola. The xx-coordinate of the vertex is h=b2a=42(1)=2h = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2. Substituting x=2x = 2 into the function gives the yy-coordinate: k=f(2)=224(2)5=9k = f(2) = 2^2 - 4(2) - 5 = -9. The height of the triangle is the vertical distance from the xx-axis to the vertex, which is 9=9|-9| = 9. The area of the triangle is 12×base×height=12×6×9=27\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 9 = 27.

Adım Adım Çözüm

1
Find the xx-intercepts of the parabola.
The intercepts are (1,0)(-1, 0) and (5,0)(5, 0).
Setting f(x)=0f(x) = 0 gives x24x5=0x^2 - 4x - 5 = 0. Factoring the quadratic equation yields (x5)(x+1)=0(x - 5)(x + 1) = 0, which gives x=5x = 5 and x=1x = -1.
2
Calculate the base of the triangle.
The base length is 66.
The base of the triangle is the segment along the xx-axis between the two intercepts. The distance between (1,0)(-1, 0) and (5,0)(5, 0) is 5(1)=65 - (-1) = 6.
3
Find the vertex (h,k)(h, k) of the parabola.
The vertex is at (2,9)(2, -9).
The xx-coordinate of the vertex is the midpoint of the intercepts: h=1+52=2h = \frac{-1 + 5}{2} = 2. The yy-coordinate is k=f(2)=224(2)5=485=9k = f(2) = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9.
4
Calculate the area of the triangle.
The area is 2727.
The height of the triangle is the distance from the xx-axis to the vertex, which is k=9=9|k| = |-9| = 9. Using the formula for the area of a triangle, Area=12×base×height=12×6×9=27\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 9 = 27.

Anahtar Kavram

Finding the xx-intercepts and vertex of a quadratic function to solve geometric problems in the coordinate plane.
Soru 293Soru

The function ff is defined by f(x)=(x3)(x3kx2+5x15)f(x) = (x - 3)(x^3 - kx^2 + 5x - 15), where kk is a constant. In the xyxy-plane, the graph of y=f(x)y = f(x) is tangent to the xx-axis at the point (3,0)(3, 0). What is the value of kk?

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Cevap: 3

Cevap

3
For the graph of a polynomial function to be tangent to the xx-axis at (3,0)(3, 0), the root x=3x = 3 must have an even multiplicity (at least 2). The function is defined as f(x)=(x3)(x3kx2+5x15)f(x) = (x - 3)(x^3 - kx^2 + 5x - 15). Since there is already one factor of (x3)(x - 3) explicitly defined, the remaining cubic factor g(x)=x3kx2+5x15g(x) = x^3 - kx^2 + 5x - 15 must also have a factor of (x3)(x - 3) to make the total multiplicity of the root x=3x = 3 at least 2. According to the Factor Theorem, if (x3)(x - 3) is a factor of g(x)g(x), then g(3)=0g(3) = 0. Substituting x=3x = 3 into g(x)g(x) gives 33k(3)2+5(3)15=03^3 - k(3)^2 + 5(3) - 15 = 0, which simplifies to 279k=027 - 9k = 0. Solving this equation for kk yields k=3k = 3.

Adım Adım Çözüm

1
Identify the relationship between graph tangency and factor multiplicity.
For the graph of a polynomial function to be tangent to the xx-axis at a point (c,0)(c, 0), the factor (xc)(x - c) must have an even multiplicity of at least 2 in the polynomial's factored form.
An odd multiplicity root causes the graph to cross the xx-axis, while an even multiplicity root causes the graph to touch the xx-axis and turn around (tangency).
2
Apply the multiplicity requirement to the given function.
Since f(x)=(x3)(x3kx2+5x15)f(x) = (x - 3)(x^3 - kx^2 + 5x - 15) already contains one factor of (x3)(x - 3), the cubic expression g(x)=x3kx2+5x15g(x) = x^3 - kx^2 + 5x - 15 must also contain (x3)(x - 3) as a factor to ensure the total multiplicity of the root x=3x = 3 is at least 2.
This guarantees that (x3)2(x - 3)^2 is a factor of f(x)f(x).
3
Apply the Factor Theorem to the cubic expression.
Since (x3)(x - 3) is a factor of g(x)g(x), then g(3)=0g(3) = 0.
The Factor Theorem states that a polynomial P(x)P(x) has a factor (xc)(x - c) if and only if P(c)=0P(c) = 0.
4
Solve for the constant kk by substituting x=3x = 3 into g(x)g(x).
33k(3)2+5(3)15=0    279k+1515=0    279k=0    9k=27    k=33^3 - k(3)^2 + 5(3) - 15 = 0 \implies 27 - 9k + 15 - 15 = 0 \implies 27 - 9k = 0 \implies 9k = 27 \implies k = 3.
Arithmetic simplification yields the value of the constant.

Anahtar Kavram

The relationship between polynomial factors, root multiplicities, and the behavior of the graph at xx-intercepts.
Soru 294Soru

A polynomial function qq with real coefficients satisfies the equation q(x)+q(6x)=8q(x) + q(6-x) = 8 for all real numbers xx. In the xyxy-plane, the graph of y=q(x)y = q(x) has an xx-intercept at (5,0)(5, 0). What is the remainder when q(x)q(x) is divided by x1x-1?

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Cevap: 88

Cevap

The correct answer is 88, which is the remainder when the polynomial is divided by x1x-1.
To find the remainder when the polynomial q(x)q(x) is divided by x1x-1, we use the Remainder Theorem, which states that this remainder is equal to q(1)q(1). Since the graph of y=q(x)y = q(x) has an xx-intercept at (5,0)(5, 0), we know that q(5)=0q(5) = 0. Substituting x=5x = 5 into the given equation q(x)+q(6x)=8q(x) + q(6-x) = 8 yields q(5)+q(65)=8q(5) + q(6-5) = 8, which simplifies to q(5)+q(1)=8q(5) + q(1) = 8. Substituting q(5)=0q(5) = 0 into this equation gives 0+q(1)=80 + q(1) = 8, so q(1)=8q(1) = 8. Geometrically, the equation q(x)+q(6x)2=4\frac{q(x) + q(6-x)}{2} = 4 shows that the graph of qq is symmetric under a point reflection about (3,4)(3, 4). Reflecting the point (5,0)(5, 0) about (3,4)(3, 4) gives the point (1,8)(1, 8), verifying that q(1)=8q(1) = 8.

Adım Adım Çözüm

1
Translate the given graphical intercept into function notation.
q(5)=0q(5) = 0
An xx-intercept at (5,0)(5, 0) means that when the input xx is 55, the output of the function is 00.
2
Apply the Remainder Theorem to the divisor x1x-1.
The remainder is equal to q(1)q(1).
The Remainder Theorem states that when a polynomial q(x)q(x) is divided by xax-a, the remainder is q(a)q(a).
3
Substitute x=5x = 5 into the given functional equation q(x)+q(6x)=8q(x) + q(6-x) = 8.
q(5)+q(65)=8    q(5)+q(1)=8q(5) + q(6-5) = 8 \implies q(5) + q(1) = 8
Substituting x=5x = 5 allows us to relate the known value q(5)q(5) directly to the target value q(1)q(1).
4
Solve for q(1)q(1) using the value of q(5)q(5).
0+q(1)=8    q(1)=80 + q(1) = 8 \implies q(1) = 8
Substituting q(5)=0q(5) = 0 into the equation yields the value of q(1)q(1), which is the remainder.

Anahtar Kavram

Remainder Theorem and Polynomial Graphs

Alternatif Yöntem

Use the point symmetry of the graph. The equation q(x)+q(6x)=8q(x) + q(6-x) = 8 can be rewritten as q(x)+q(6x)2=4\frac{q(x) + q(6-x)}{2} = 4. This indicates that the graph of y=q(x)y = q(x) has point symmetry about the point (3,4)(3, 4). Since the graph has an xx-intercept at (5,0)(5, 0), point symmetry dictates that the reflected point across (3,4)(3, 4) must also lie on the graph. The midpoint of (5,0)(5, 0) and the symmetric point (x0,y0)(x_0, y_0) is (3,4)(3, 4). Solving 5+x02=3\frac{5+x_0}{2} = 3 gives x0=1x_0 = 1, and 0+y02=4\frac{0+y_0}{2} = 4 gives y0=8y_0 = 8. Thus, (1,8)(1, 8) lies on the graph, meaning q(1)=8q(1) = 8. By the Remainder Theorem, the remainder when q(x)q(x) is divided by x1x-1 is q(1)=8q(1) = 8.
Tahmini Süre:1m 30s
Soru 295Soru
An equation is shown below.
xx12x+2=6x2+x2\frac{x}{x - 1} - \frac{2}{x + 2} = \frac{6}{x^2 + x - 2}
What is the value of the real solution to the equation?
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Cevap: 2

Cevap

The correct answer is 2.
The correct answer is 2. Multiplying both sides by the least common denominator (x1)(x+2)(x - 1)(x + 2) results in the quadratic equation x(x+2)2(x1)=6x(x + 2) - 2(x - 1) = 6. Simplifying this equation yields x2+2=6x^2 + 2 = 6, which has solutions x=2x = 2 and x=2x = -2. However, substituting x=2x = -2 into the original equation results in division by zero, making it an extraneous solution. Therefore, 22 is the only valid real solution.

Adım Adım Çözüm

1
Multiply the entire equation by the least common denominator, (x1)(x+2)=x2+x2(x - 1)(x + 2) = x^2 + x - 2, to clear the denominators.
x(x+2)2(x1)=6x(x + 2) - 2(x - 1) = 6
This simplifies the rational equation into a polynomial equation.
2
Expand the terms and simplify the equation.
x2+2x2x+2=6x^2 + 2x - 2x + 2 = 6, which simplifies to x2+2=6x^2 + 2 = 6.
Distributing the terms allows us to group like terms and solve for the variable.
3
Solve the quadratic equation for xx.
x2=4x^2 = 4, which gives x=2x = 2 or x=2x = -2.
Subtracting 2 from both sides isolates the squared variable.
4
Check for extraneous solutions by substituting the potential solutions back into the original denominators.
For x=2x = -2, the denominator x+2x + 2 becomes 0, which is undefined. For x=2x = 2, all denominators are non-zero.
Solutions that make any denominator in the original equation equal to zero are extraneous and must be excluded.

Anahtar Kavram

Solving rational equations by finding a common denominator and checking for extraneous solutions.
Soru 296Soru

For the quadratic function ff, the table below shows three points that lie on its graph in the xyxy-plane, where kk is a constant.

xxf(x)f(x)
2200
6600
1115-15

If the vertex of the graph of y=f(x)y = f(x) is (4,k)(4, k), what is the value of kk?

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Cevap: 12

Cevap

The value of kk is 1212.
The table indicates that the function has xx-intercepts at x=2x = 2 and x=6x = 6 because f(2)=0f(2) = 0 and f(6)=0f(6) = 0. Consequently, the quadratic function can be written in factored form as f(x)=a(x2)(x6)f(x) = a(x - 2)(x - 6) for some constant aa. Using the point (1,15)(1, -15) to find aa, we substitute x=1x = 1 and f(1)=15f(1) = -15, which yields 15=a(12)(16)-15 = a(1 - 2)(1 - 6), or 15=5a-15 = 5a, so a=3a = -3. Since the vertex of the graph is (4,k)(4, k), the value of kk is the function value at the vertex's xx-coordinate, which is f(4)f(4). Evaluating the function gives k=3(42)(46)=3(2)(2)=12k = -3(4 - 2)(4 - 6) = -3(2)(-2) = 12.

Adım Adım Çözüm

1
Write the quadratic function in factored form using the given xx-intercepts.
f(x)=a(x2)(x6)f(x) = a(x - 2)(x - 6)
The table shows that f(2)=0f(2) = 0 and f(6)=0f(6) = 0, meaning the graph has xx-intercepts at x=2x = 2 and x=6x = 6.
2
Substitute the point (1,15)(1, -15) into the factored equation to find the value of the constant aa.
a=3a = -3
Substituting x=1x = 1 and f(x)=15f(x) = -15 gives 15=a(12)(16)    15=5a    a=3-15 = a(1 - 2)(1 - 6) \implies -15 = 5a \implies a = -3.
3
Calculate the value of kk by finding the function value at the vertex x=4x = 4.
k=12k = 12
Since the vertex is (4,k)(4, k), the value of kk is f(4)f(4). Substituting x=4x = 4 into f(x)=3(x2)(x6)f(x) = -3(x - 2)(x - 6) yields k=3(42)(46)=3(2)(2)=12k = -3(4 - 2)(4 - 6) = -3(2)(-2) = 12.

Anahtar Kavram

Using intercepts and an additional point to determine the equation of a quadratic function, and evaluating it at the vertex.
Soru 297Soru

The function ff is defined by f(x)=2x3f(x) = 2x - 3. If the function gg is defined by g(x)=f(x+4)g(x) = f(x + 4), what is the value of g(1)g(1)?

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Cevap: 7

Cevap

The correct answer is 7.
Since g(x)=f(x+4)g(x) = f(x + 4), evaluating g(1)g(1) requires finding f(1+4)f(1 + 4), which is f(5)f(5). Substituting 55 into the expression for f(x)f(x) gives f(5)=2(5)3=103=7f(5) = 2(5) - 3 = 10 - 3 = 7. Therefore, the correct answer is 7.

Adım Adım Çözüm

1
Substitute x=1x = 1 into the definition of g(x)g(x) to express g(1)g(1) in terms of ff.
g(1)=f(1+4)=f(5)g(1) = f(1 + 4) = f(5)
To find the value of g(1)g(1), we must evaluate the input to the outer function first.
2
Substitute the input value 55 into the expression for f(x)f(x).
f(5)=2(5)3f(5) = 2(5) - 3
Evaluating f(5)f(5) requires replacing xx with 55 in the definition f(x)=2x3f(x) = 2x - 3.
3
Perform the operations to find the final value.
f(5)=103=7f(5) = 10 - 3 = 7
Simplifying the numerical expression gives the final value of g(1)g(1).

Anahtar Kavram

Evaluating a transformed function at a given point using function notation.
Tahmini Süre:45s
Soru 298Soru

A projectile is launched from the ground. Its height, in feet, tt seconds after launch is modeled by the function h(t)=16t2+v0th(t) = -16t^2 + v_0 t, where v0v_0 is the initial upward velocity in feet per second. If the projectile reaches its maximum height of 144144 feet, what is the value of v0v_0?

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Cevap: 96

Cevap

The correct answer is 96. The initial velocity of the projectile must be 96 feet per second to reach a maximum height of 144 feet.
The maximum height of a projectile modeled by a quadratic function is the yy-value of its vertex. For h(t)=16t2+v0th(t) = -16t^2 + v_0 t, the time tt at the vertex is given by t=b2a=v02(16)=v032t = -\frac{b}{2a} = -\frac{v_0}{2(-16)} = \frac{v_0}{32}. Substituting this value of tt back into the height equation yields the maximum height: h(v032)=16(v032)2+v0(v032)=16(v021024)+v0232=v0264+v0232=v0264h\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0^2}{1024}\right) + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{v_0^2}{64}. Given that the maximum height is 144144 feet, we set v0264=144\frac{v_0^2}{64} = 144, which simplifies to v02=144×64v_0^2 = 144 \times 64. Taking the square root of both sides gives v0=12×8=96v_0 = 12 \times 8 = 96.

Adım Adım Çözüm

1
Find the time tt at which the maximum height occurs using the vertex formula.
t=v032t = \frac{v_0}{32}
For a quadratic function of the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the vertex occurs at x=b2ax = -\frac{b}{2a}. Here, a=16a = -16 and b=v0b = v_0, so the vertex is at t=v02(16)=v032t = -\frac{v_0}{2(-16)} = \frac{v_0}{32}.
2
Substitute the time t=v032t = \frac{v_0}{32} into the height function to express the maximum height in terms of v0v_0.
h(v032)=v0264h\left(\frac{v_0}{32}\right) = \frac{v_0^2}{64}
Evaluating h(v032)h\left(\frac{v_0}{32}\right) gives 16(v032)2+v0(v032)=16(v021024)+v0232=v0264+v0232=v0264-16\left(\frac{v_0}{32}\right)^2 + v_0\left(\frac{v_0}{32}\right) = -16\left(\frac{v_0^2}{1024}\right) + \frac{v_0^2}{32} = -\frac{v_0^2}{64} + \frac{v_0^2}{32} = \frac{v_0^2}{64}.
3
Set the maximum height expression equal to the given maximum height of 144144 feet and solve for v0v_0.
v0=96v_0 = 96
Setting v0264=144\frac{v_0^2}{64} = 144 yields v02=144×64=9216v_0^2 = 144 \times 64 = 9216. Taking the positive square root (since v0v_0 represents a positive initial upward velocity) gives v0=144×64=12×8=96v_0 = \sqrt{144} \times \sqrt{64} = 12 \times 8 = 96.

Anahtar Kavram

Determining the vertex coordinates of a quadratic function to find maximum value in context.
Soru 299Soru

If 272x2=(13)x827^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}, what is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of xx is 22.
To solve the equation 272x2=(13)x827^{2x - 2} = \left(\frac{1}{3}\right)^{x - 8}, express both sides with the base 33. Since 27=3327 = 3^3 and 13=31\frac{1}{3} = 3^{-1}, the equation becomes (33)2x2=(31)x8(3^3)^{2x - 2} = (3^{-1})^{x - 8}. Applying the exponent rule (am)n=amn(a^m)^n = a^{mn} yields 36x6=38x3^{6x - 6} = 3^{8 - x}. Since the bases are now the same, set the exponents equal to each other: 6x6=8x6x - 6 = 8 - x. Adding xx to both sides gives 7x6=87x - 6 = 8, and adding 66 to both sides gives 7x=147x = 14. Dividing by 77 results in x=2x = 2.

Adım Adım Çözüm

1
Rewrite 2727 and 13\frac{1}{3} as powers of 33.
27=3327 = 3^3 and 13=31\frac{1}{3} = 3^{-1}
This allows both sides of the equation to be expressed with a common base of 33 so their exponents can be equated.
2
Substitute these bases back into the equation and apply exponent rules.
(33)2x2=(31)x8    33(2x2)=3(x8)    36x6=38x(3^3)^{2x - 2} = (3^{-1})^{x - 8} \implies 3^{3(2x - 2)} = 3^{-(x - 8)} \implies 3^{6x - 6} = 3^{8 - x}
Using the power rule (am)n=amn(a^m)^n = a^{mn}, multiply the outer exponent by the power of the base.
3
Equate the exponents and solve for xx.
6x6=8x    7x=14    x=26x - 6 = 8 - x \implies 7x = 14 \implies x = 2
Since the bases are equal, their exponents must be equal for the equation to hold true.

Anahtar Kavram

Solving exponential equations by finding a common base and applying exponent rules.
Soru 300Soru

The quadratic function ff is defined by f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where aa, hh, and kk are constants. In the xyxy-plane, the graph of y=f(x)y = f(x) has a vertex at (3,4)(3, -4) and passes through the point (5,8)(5, 8). If the function gg is defined by g(x)=2f(x1)+5g(x) = -2f(x - 1) + 5, what is the value of g(2)g(2)?

Cevabı ve açıklamayı göster

Cevap: -11

Cevap

-11
To find the value of g(2)g(2), we first determine the equation of the quadratic function f(x)f(x). Since the vertex is (3,4)(3, -4), the vertex form is f(x)=a(x3)24f(x) = a(x - 3)^2 - 4. Substituting the point (5,8)(5, 8) yields 8=a(53)248 = a(5 - 3)^2 - 4, which simplifies to 12=4a12 = 4a, so a=3a = 3. Therefore, f(x)=3(x3)24f(x) = 3(x - 3)^2 - 4. We then substitute x=2x = 2 into the definition of g(x)g(x), obtaining g(2)=2f(21)+5=2f(1)+5g(2) = -2f(2 - 1) + 5 = -2f(1) + 5. Evaluating f(1)f(1) gives f(1)=3(13)24=8f(1) = 3(1 - 3)^2 - 4 = 8. Substituting this back into the expression for g(2)g(2) gives 2(8)+5=11-2(8) + 5 = -11. Thus, the option with value -11 is correct.

Adım Adım Çözüm

1
Write the quadratic function f(x)f(x) in vertex form using the given vertex (3,4)(3, -4).
f(x)=a(x3)24f(x) = a(x - 3)^2 - 4
The vertex form of a quadratic function is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
2
Substitute the point (5,8)(5, 8) into the vertex form to solve for the constant aa.
8=a(53)24    8=4a4    12=4a    a=38 = a(5 - 3)^2 - 4 \implies 8 = 4a - 4 \implies 12 = 4a \implies a = 3. Thus, f(x)=3(x3)24f(x) = 3(x - 3)^2 - 4.
Since the graph of ff passes through (5,8)(5, 8), these coordinates must satisfy the function's equation.
3
Substitute x=2x = 2 into the definition of g(x)g(x) to express g(2)g(2) in terms of ff.
g(2)=2f(21)+5=2f(1)+5g(2) = -2f(2 - 1) + 5 = -2f(1) + 5
We need to evaluate the inner function transformation f(x1)f(x - 1) at x=2x = 2.
4
Evaluate f(1)f(1) using the formula determined in Step 2.
f(1)=3(13)24=3(2)24=3(4)4=8f(1) = 3(1 - 3)^2 - 4 = 3(-2)^2 - 4 = 3(4) - 4 = 8
To find g(2)g(2), we must first compute the value of f(1)f(1).
5
Substitute f(1)=8f(1) = 8 back into the expression for g(2)g(2) and simplify.
g(2)=2(8)+5=16+5=11g(2) = -2(8) + 5 = -16 + 5 = -11
This completes the evaluation of the multi-step transformation.

Anahtar Kavram

Function Notation and Transformations
Tahmini Süre:2m 0s
ÖncekiSayfa 15 / 22Sonraki
Advanced Math Alıştırma Soruları — SAT — Sayfa 15 | Examkin