Advanced Math

438 soru

Soru 381Soru

A transformation is applied to the exponential function f(x)=2x3f(x) = 2^x - 3 to obtain a new function gg, where g(x)=f(x1)+2g(x) = -f(x - 1) + 2. What is the yy-intercept of the graph of y=g(x)y = g(x) in the xyxy-plane?

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Cevap: (0,92)(0, \frac{9}{2})

Cevap

(0,92)(0, \frac{9}{2})
The correct answer is the coordinate point with yy-coordinate 92\frac{9}{2}. The yy-intercept of the graph of g(x)g(x) is found by evaluating g(0)g(0). Substituting x=0x = 0 into the transformation equation gives g(0)=f(1)+2g(0) = -f(-1) + 2. Using the definition of f(x)f(x), we find f(1)=213=0.53=2.5f(-1) = 2^{-1} - 3 = 0.5 - 3 = -2.5. Substituting this value back into the equation for g(0)g(0) yields g(0)=(2.5)+2=2.5+2=4.5g(0) = -(-2.5) + 2 = 2.5 + 2 = 4.5, which is equivalent to 92\frac{9}{2}. Therefore, the yy-intercept is the coordinate point containing 92\frac{9}{2}.

Adım Adım Çözüm

1
Substitute x=0x = 0 into the definition of g(x)g(x) to find the yy-coordinate of the yy-intercept.
g(0)=f(01)+2=f(1)+2g(0) = -f(0 - 1) + 2 = -f(-1) + 2
The yy-intercept of any function y=g(x)y = g(x) occurs where the input xx is equal to 00.
2
Evaluate f(1)f(-1) using the definition f(x)=2x3f(x) = 2^x - 3.
f(1)=213=123=2.5f(-1) = 2^{-1} - 3 = \frac{1}{2} - 3 = -2.5
Evaluating f(1)f(-1) is necessary to substitute its value back into the expression for g(0)g(0).
3
Substitute the value of f(1)f(-1) back into the expression for g(0)g(0) and simplify.
g(0)=(2.5)+2=2.5+2=4.5=92g(0) = -(-2.5) + 2 = 2.5 + 2 = 4.5 = \frac{9}{2}
This yields the final yy-coordinate of the yy-intercept, which is 92\frac{9}{2}, giving the coordinate point (0,92)(0, \frac{9}{2}).

Anahtar Kavram

Applying horizontal translation, reflection across the x-axis, and vertical translation to evaluate a function at a specific point.
Soru 382Soru

The height of a diver, in meters, above the pool surface tt seconds after leaving the diving board is modeled by the function f(t)=4.9(t1)2+10f(t) = -4.9(t - 1)^2 + 10. If the diving board is moved so that the diver's trajectory is shifted 0.50.5 seconds later in time and the maximum height is increased by 22 meters, which of the following functions gg models the diver's new trajectory?

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Cevap: g(t)=4.9(t1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12

Cevap

The function g(t)=4.9(t1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12 models the diver's new trajectory.
The correct answer represents the translated quadratic function. Since the original function f(t)=4.9(t1)2+10f(t) = -4.9(t - 1)^2 + 10 has its vertex at (1,10)(1, 10), shifting the trajectory 0.50.5 seconds later in time moves the vertex horizontally to the right to t=1.5t = 1.5 seconds, replacing (t1)(t - 1) with (t1.5)(t - 1.5). Increasing the maximum height by 22 meters shifts the vertex vertically upward to 1212 meters, replacing the constant term 1010 with 1212. This results in the function g(t)=4.9(t1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12.

Adım Adım Çözüm

1
Identify the vertex and its meaning in the original function f(t)=4.9(t1)2+10f(t) = -4.9(t - 1)^2 + 10.
The vertex is (1,10)(1, 10), indicating that the maximum height of 1010 meters occurs at t=1t = 1 second.
The vertex form of a quadratic function is y=a(th)2+ky = a(t - h)^2 + k, where (h,k)(h, k) is the vertex representing the extreme value.
2
Apply the horizontal translation of 0.50.5 seconds later in time.
The new tt-coordinate of the vertex is 1+0.5=1.51 + 0.5 = 1.5 seconds, which changes the term (t1)(t - 1) to (t1.5)(t - 1.5).
A horizontal shift to the right by cc units is represented by replacing the variable tt with tct - c.
3
Apply the vertical translation of 22 meters upward.
The new yy-coordinate of the vertex is 10+2=1210 + 2 = 12 meters, which changes the constant term of the function to 1212.
A vertical shift upward by dd units is represented by adding dd to the function value.
4
Combine the horizontal and vertical translations to write the new equation.
g(t)=4.9(t1.5)2+12g(t) = -4.9(t - 1.5)^2 + 12.
Substituting the shifted vertex coordinates (1.5,12)(1.5, 12) into the vertex form while keeping the same vertical stretch/direction factor a=4.9a = -4.9 yields the final equation.

Anahtar Kavram

Vertex form and transformations of quadratic functions
Soru 383Soru

The function ff is defined by f(x)=2x212x+cf(x) = 2x^2 - 12x + c, where cc is a constant. In the xyxy-plane, the graph of y=f(x)y = f(x) has a vertex at (h,5)(h, 5), where hh is a constant. What is the value of cc?

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Cevap: 23

Cevap

23
For a quadratic function in standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the x-coordinate of the vertex is given by h=b2ah = -\frac{b}{2a}. In this function, a=2a = 2 and b=12b = -12, so h=122(2)=3h = -\frac{-12}{2(2)} = 3. The vertex coordinates are (3,5)(3, 5), which means f(3)=5f(3) = 5. Substituting x=3x = 3 into the function gives 2(3)212(3)+c=52(3)^2 - 12(3) + c = 5. Simplifying the terms gives 1836+c=518 - 36 + c = 5, which becomes 18+c=5-18 + c = 5. Adding 1818 to both sides yields c=23c = 23.

Adım Adım Çözüm

1
Identify the x-coordinate formula for the vertex of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c.
The x-coordinate hh is given by h=b2ah = -\frac{b}{2a}.
To find the axis of symmetry and the horizontal position of the vertex.
2
Substitute a=2a = 2 and b=12b = -12 from the given equation f(x)=2x212x+cf(x) = 2x^2 - 12x + c into the vertex formula.
h=122(2)=3h = -\frac{-12}{2(2)} = 3
To calculate the specific x-coordinate of the vertex for this function.
3
Substitute the vertex coordinates (3,5)(3, 5) into the function f(x)f(x).
f(3)=2(3)212(3)+c=5f(3) = 2(3)^2 - 12(3) + c = 5
Since the vertex lies on the graph of the function, its coordinates must satisfy the function's equation.
4
Simplify the equation and solve for the constant cc.
1836+c=5    18+c=5    c=2318 - 36 + c = 5 \implies -18 + c = 5 \implies c = 23
To determine the final value of the constant cc.

Anahtar Kavram

Vertex of a quadratic function
Soru 384Soru

The function ff is defined by f(x)=x23f(x) = |x - 2| - 3. The function gg is defined by g(x)=f(x+1)+2g(x) = -f(x + 1) + 2. What is the maximum value of g(x)g(x)?

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Cevap: 5

Cevap

5
The vertex of the absolute value function f(x)=x23f(x) = |x - 2| - 3 is at (2,3)(2, -3), meaning its minimum value is 3-3. The transformed function g(x)=f(x+1)+2g(x) = -f(x + 1) + 2 shifts the graph 1 unit to the left, reflects it vertically across the xx-axis (which turns the minimum value of 3-3 into a maximum value of 33), and translates it 2 units upward. This results in a final maximum value of 3+2=53 + 2 = 5.

Adım Adım Çözüm

1
Find the minimum value of f(x)f(x) and the location of its vertex.
The vertex of f(x)=x23f(x) = |x - 2| - 3 is at (2,3)(2, -3), which means f(x)f(x) has a minimum value of 3-3 at x=2x = 2.
Since the coefficient of the absolute value expression is positive, the graph of ff opens upward, making the vertex a minimum point.
2
Apply the horizontal translation to find the new vertex position.
The term f(x+1)f(x + 1) shifts the graph 1 unit to the left, moving the vertex from x=2x = 2 to x=1x = 1.
Adding a constant inside the function argument, f(x+h)f(x + h) where h>0h > 0, shifts the graph horizontally to the left by hh units.
3
Apply the vertical reflection to the minimum value.
The term f(x+1)-f(x + 1) reflects the graph across the xx-axis. The vertex point (1,3)(1, -3) becomes (1,3)(1, 3), and the minimum value of 3-3 becomes a maximum value of 33.
Multiplying the function by 1-1 reflects the graph vertically, swapping all positive and negative yy-values and changing the minimum value to a maximum value.
4
Apply the vertical translation to find the final maximum value.
Adding 2 to the function, g(x)=f(x+1)+2g(x) = -f(x + 1) + 2, shifts the entire graph vertically upward by 2 units. The vertex moves from (1,3)(1, 3) to (1,5)(1, 5).
Adding a constant to the outside of the function shifts the graph vertically, so the maximum value increases from 3 to 5.

Anahtar Kavram

Function Notation and Transformations
Soru 385Soru

The graph of the quadratic function f(x)=x24x+7f(x) = x^2 - 4x + 7 is translated 33 units to the right and 22 units down in the xyxy-plane to form the graph of the function g(x)=x2+px+qg(x) = x^2 + px + q, where pp and qq are constants. What is the value of qq?

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Cevap: 26

Cevap

The value of qq is 2626.
To find the constant term qq of the translated quadratic function g(x)=x2+px+qg(x) = x^2 + px + q, we first determine the vertex of the original function f(x)=x24x+7f(x) = x^2 - 4x + 7. Completing the square gives f(x)=(x2)2+3f(x) = (x - 2)^2 + 3, which identifies the vertex of the parabola as (2,3)(2, 3). Translating the graph 33 units to the right and 22 units down shifts the vertex to (2+3,32)=(5,1)(2 + 3, 3 - 2) = (5, 1). Because the translation does not affect the shape of the parabola, the leading coefficient remains 11. The vertex form of the new function is g(x)=(x5)2+1g(x) = (x - 5)^2 + 1. Expanding this expression yields g(x)=x210x+25+1=x210x+26g(x) = x^2 - 10x + 25 + 1 = x^2 - 10x + 26. Comparing this to g(x)=x2+px+qg(x) = x^2 + px + q, we find that q=26q = 26.

Adım Adım Çözüm

1
Convert the original function f(x)=x24x+7f(x) = x^2 - 4x + 7 into vertex form, f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, to identify its vertex (h,k)(h, k).
f(x)=(x2)2+3f(x) = (x - 2)^2 + 3, which represents a parabola with vertex (2,3)(2, 3).
Finding the vertex of the original function allows us to apply the translation directly to the vertex coordinates.
2
Apply the translation of 33 units to the right and 22 units down to the coordinates of the vertex (2,3)(2, 3).
The new vertex is (2+3,32)=(5,1)(2 + 3, 3 - 2) = (5, 1).
Translating a graph shifts its vertex by the corresponding horizontal and vertical amounts.
3
Write the equation of the translated function g(x)g(x) in vertex form using the new vertex (5,1)(5, 1) and the original leading coefficient a=1a = 1.
g(x)=(x5)2+1g(x) = (x - 5)^2 + 1
A translation does not change the shape or vertical stretch of the parabola, so the coefficient of x2x^2 remains 11.
4
Expand the vertex form of g(x)g(x) into standard form, g(x)=x2+px+qg(x) = x^2 + px + q, to determine the constant term qq.
g(x)=x210x+26g(x) = x^2 - 10x + 26, which means q=26q = 26.
Expanding the equation allows us to compare it directly with the standard form of g(x)g(x) and identify the value of the constant term.

Anahtar Kavram

Vertex form and translations of quadratic functions

Alternatif Yöntem

Alternatively, the translation can be applied directly to the variable xx in the function equation. Translating a function f(x)f(x) by 33 units to the right and 22 units down yields g(x)=f(x3)2g(x) = f(x - 3) - 2. Substituting x3x - 3 into the original function gives: g(x)=(x3)24(x3)+72g(x) = (x - 3)^2 - 4(x - 3) + 7 - 2. Simplifying this expression: g(x)=(x26x+9)(4x12)+5=x210x+26g(x) = (x^2 - 6x + 9) - (4x - 12) + 5 = x^2 - 10x + 26. This directly shows that the constant term qq is 2626.
Tahmini Süre:1m 30s
Soru 386Soru

A company's daily profit, P(x)P(x), in dollars, from selling xx units of a product is modeled by the function P(x)=2x2+kx800P(x) = -2x^2 + kx - 800, where kk is a constant. If the company achieves its maximum daily profit of $1000\$1000 when it sells 3030 units, what is the value of kk?

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Cevap: 120

Cevap

120
The maximum value of a quadratic function occurs at its vertex. Given that the vertex is at (30,1000)(30, 1000) and the coefficient of the squared term is 2-2, the profit function can be written in vertex form as P(x)=2(x30)2+1000P(x) = -2(x - 30)^2 + 1000. Expanding this expression yields P(x)=2(x260x+900)+1000=2x2+120x800P(x) = -2(x^2 - 60x + 900) + 1000 = -2x^2 + 120x - 800. Comparing this to the given equation P(x)=2x2+kx800P(x) = -2x^2 + kx - 800, the coefficient of xx must be 120. Alternatively, using the vertex formula h=b/(2a)h = -b/(2a) with h=30h = 30 and a=2a = -2 gives 30=k/(2×2)30 = -k/(2 \times -2), which simplifies to 30=k/430 = k/4 and results in k=120k = 120.

Adım Adım Çözüm

1
Identify the vertex coordinates from the given context.
The vertex of the profit parabola is at (h,d)=(30,1000)(h, d) = (30, 1000).
The maximum profit of $1000\$1000 occurs when 3030 units are sold, representing the peak of the downward-opening parabola.
2
Substitute the vertex and the leading coefficient a=2a = -2 into the vertex form of a quadratic function, P(x)=a(xh)2+dP(x) = a(x - h)^2 + d.
The equation becomes P(x)=2(x30)2+1000P(x) = -2(x - 30)^2 + 1000.
The vertex form allows direct substitution of the vertex coordinates to build the function's equation.
3
Expand the vertex form equation into standard form.
P(x)=2(x260x+900)+1000=2x2+120x800P(x) = -2(x^2 - 60x + 900) + 1000 = -2x^2 + 120x - 800.
Expanding the equation allows direct comparison of terms with the given standard form P(x)=2x2+kx800P(x) = -2x^2 + kx - 800.
4
Compare the coefficient of the xx term in the expanded equation to the coefficient of the xx term in the given equation.
k=120k = 120.
Corresponding coefficients of identical functions must be equal, allowing us to determine the value of the constant kk.

Anahtar Kavram

Quadratic Functions and Graphs
Soru 387Soru

In the quadratic equation x2kx+36=0x^2 - kx + 36 = 0, kk is a positive constant. If the difference between the two solutions to the equation is 55, what is the value of kk?

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Cevap: 13

Cevap

The value of the positive constant kk is 1313.
For the quadratic equation x2kx+36=0x^2 - kx + 36 = 0 with solutions r1r_1 and r2r_2, the sum of the solutions is r1+r2=kr_1 + r_2 = k and the product of the solutions is r1r2=36r_1 \cdot r_2 = 36. Given that the difference between the two solutions is 55, we can write r1r2=5|r_1 - r_2| = 5. Squaring both sides yields (r1r2)2=25(r_1 - r_2)^2 = 25. Using the algebraic identity (r1r2)2=(r1+r2)24r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2, we substitute the known values to obtain 25=k24(36)25 = k^2 - 4(36), which simplifies to 25=k214425 = k^2 - 144. Solving for k2k^2 gives k2=169k^2 = 169. Since kk is positive, k=13k = 13. Alternatively, we can find two numbers whose product is 3636 and whose difference is 55. These numbers are 99 and 44, because 94=369 \cdot 4 = 36 and 94=59 - 4 = 5. The sum of these solutions is 9+4=139 + 4 = 13, which matches the coefficient of the linear term, kk.

Adım Adım Çözüm

1
Relate the roots of the quadratic equation x2kx+36=0x^2 - kx + 36 = 0 to its coefficients using Vieta's formulas.
The sum of the roots is r1+r2=kr_1 + r_2 = k and the product of the roots is r1r2=36r_1 \cdot r_2 = 36.
This sets up the system of equations representing the roots.
2
Express the given root difference of 55 mathematically and square it.
r1r2=5    (r1r2)2=25|r_1 - r_2| = 5 \implies (r_1 - r_2)^2 = 25.
Squaring the difference allows us to use standard algebraic identities.
3
Apply the identity (r1r2)2=(r1+r2)24r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2 to substitute the sum and product expressions.
25=k24(36)25 = k^2 - 4(36)
This converts the relationship between the roots into an equation with the single variable kk.
4
Solve the equation for kk, selecting the positive value.
25=k2144    k2=169    k=1325 = k^2 - 144 \implies k^2 = 169 \implies k = 13 (since k>0k > 0).
To determine the final value of kk satisfying the constraint that kk is a positive constant.

Anahtar Kavram

Relationship between the roots and coefficients of a quadratic equation (Vieta's formulas)
Soru 388Soru

A quadratic function ff has its vertex at (4,12)(4, 12) and a yy-intercept at (0,4)(0, -4) in the xyxy-plane. The function gg is defined by g(x)=f(x+2)+kg(x) = f(x + 2) + k, where kk is a constant. If the yy-intercept of the graph of gg is (0,15)(0, 15), what is the value of kk?

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Cevap: 7

Cevap

7
The quadratic function f(x)f(x) is determined to be f(x)=(x4)2+12f(x) = -(x - 4)^2 + 12 by substituting the vertex (4,12)(4, 12) and the yy-intercept (0,4)(0, -4) into the vertex form. The transformation g(x)=f(x+2)+kg(x) = f(x + 2) + k translates the function horizontally left by 2 units and vertically by kk units, resulting in g(x)=(x2)2+12+kg(x) = -(x - 2)^2 + 12 + k. Using the yy-intercept of gg, which is (0,15)(0, 15), we substitute x=0x = 0 to get 15=(02)2+12+k15 = -(0 - 2)^2 + 12 + k, simplifying to 15=8+k15 = 8 + k, which yields k=7k = 7.

Adım Adım Çözüm

1
Write the vertex form of f(x)f(x)
f(x)=a(x4)2+12f(x) = a(x - 4)^2 + 12
The vertex form of a quadratic function with vertex (h,kvertex)(h, k_{vertex}) is given by f(x)=a(xh)2+kvertexf(x) = a(x - h)^2 + k_{vertex}.
2
Determine the value of the coefficient aa
a=1a = -1, so f(x)=(x4)2+12f(x) = -(x - 4)^2 + 12
Substitute the coordinates of the yy-intercept (0,4)(0, -4) into the vertex form equation to solve for aa.
3
Express the transformed function g(x)g(x) in terms of xx and kk
g(x)=(x2)2+12+kg(x) = -(x - 2)^2 + 12 + k
Apply the translation rules: substituting x+2x + 2 for xx shifts the graph left by 2 units, and adding kk shifts the graph vertically by kk units.
4
Solve for the constant kk
k=7k = 7
Use the yy-intercept of g(x)g(x), which is (0,15)(0, 15), so g(0)=15g(0) = 15. Setting 15=(02)2+12+k15 = -(0 - 2)^2 + 12 + k simplifies to 15=8+k15 = 8 + k, which yields k=7k = 7.

Anahtar Kavram

Quadratic function vertex form and transformations
Soru 389Soru

The polynomial function ff is defined by f(x)=x32x213x+kf(x) = x^3 - 2x^2 - 13x + k, where kk is a constant. If x4x - 4 is a factor of f(x)f(x), what is the value of f(2)f(-2)?

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Cevap: 3030

Cevap

The correct value of f(2)f(-2) is 3030.
The correct answer is 3030. According to the Factor Theorem, if x4x - 4 is a factor of f(x)f(x), then f(4)=0f(4) = 0. Substituting x=4x = 4 into f(x)=x32x213x+kf(x) = x^3 - 2x^2 - 13x + k gives 432(4)213(4)+k=04^3 - 2(4)^2 - 13(4) + k = 0. Simplifying this expression results in 643252+k=064 - 32 - 52 + k = 0, which gives 20+k=0-20 + k = 0, so k=20k = 20. The polynomial is therefore f(x)=x32x213x+20f(x) = x^3 - 2x^2 - 13x + 20. Evaluating this function at x=2x = -2 gives f(2)=(2)32(2)213(2)+20=88+26+20=30f(-2) = (-2)^3 - 2(-2)^2 - 13(-2) + 20 = -8 - 8 + 26 + 20 = 30.

Adım Adım Çözüm

1
Apply the Factor Theorem to set up an equation for kk.
f(4)=0f(4) = 0
According to the Factor Theorem, if xcx - c is a factor of a polynomial f(x)f(x), then f(c)=0f(c) = 0. Since x4x - 4 is a factor, substituting x=4x = 4 into f(x)f(x) must yield 00.
2
Substitute x=4x = 4 into the polynomial and solve for kk.
k=20k = 20
Evaluating f(4)f(4) gives 432(4)213(4)+k=04^3 - 2(4)^2 - 13(4) + k = 0. This simplifies to 643252+k=064 - 32 - 52 + k = 0, which simplifies further to 20+k=0-20 + k = 0. Solving this equation gives k=20k = 20.
3
Write the complete polynomial function using the solved value of kk.
f(x)=x32x213x+20f(x) = x^3 - 2x^2 - 13x + 20
Replacing the constant kk with 2020 in the original function definition gives the complete polynomial expression.
4
Evaluate the polynomial function at x=2x = -2.
f(2)=30f(-2) = 30
Substituting x=2x = -2 into the polynomial gives f(2)=(2)32(2)213(2)+20f(-2) = (-2)^3 - 2(-2)^2 - 13(-2) + 20. Evaluating the terms gives 88+26+20=30-8 - 8 + 26 + 20 = 30.

Anahtar Kavram

The Factor Theorem states that a polynomial f(x)f(x) has a factor xcx - c if and only if f(c)=0f(c) = 0. This can be used to determine unknown coefficients in a polynomial before evaluating the function at a different value.
Soru 390Soru

In the xyxy-plane, the graph of the quadratic function ff is a parabola with vertex (4,3)(4, -3). If the graph passes through the point (1,15)(1, 15), what is the value of f(2)f(2)?

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Cevap: 5

Cevap

The value of f(2)f(2) is 55.
By using the vertex form of a quadratic function, f(x)=a(xh)2+kf(x) = a(x - h)^2 + k with vertex (4,3)(4, -3), the function can be written as f(x)=a(x4)23f(x) = a(x - 4)^2 - 3. Substituting the point (1,15)(1, 15) gives 15=a(14)2315 = a(1 - 4)^2 - 3, which simplifies to 18=9a18 = 9a, leading to a=2a = 2. Substituting a=2a = 2 back into the function gives f(x)=2(x4)23f(x) = 2(x - 4)^2 - 3. Finally, evaluating at x=2x = 2 gives f(2)=2(24)23=5f(2) = 2(2 - 4)^2 - 3 = 5.

Adım Adım Çözüm

1
Write the quadratic function in vertex form and substitute the vertex (4,3)(4, -3).
f(x)=a(x4)23f(x) = a(x - 4)^2 - 3
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 (1,15)(1, 15) into the equation and solve for the constant aa.
a=2a = 2
Since the graph passes through (1,15)(1, 15), substituting x=1x = 1 and f(x)=15f(x) = 15 allows us to solve for aa.
3
Substitute x=2x = 2 into the completed function f(x)=2(x4)23f(x) = 2(x - 4)^2 - 3 to find f(2)f(2).
f(2)=5f(2) = 5
Evaluating the function at x=2x = 2 yields the required value.

Anahtar Kavram

Determining a quadratic function's equation from its vertex and a point, then evaluating it.
Soru 391Soru

A polynomial function ff has the form f(x)=a(x2)(x+3)(x5)f(x) = a(x - 2)(x + 3)(x - 5), where aa is a constant. In the xyxy-plane, the graph of y=f(x)y = f(x) has a yy-intercept of (0,60)(0, 60). What is the value of f(1)f(1)?

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Cevap: 32

Cevap

32
The yy-intercept of the graph is given as (0,60)(0, 60), which means that when x=0x = 0, f(0)=60f(0) = 60. Substituting x=0x = 0 into the function gives f(0)=a(02)(0+3)(05)=30af(0) = a(0 - 2)(0 + 3)(0 - 5) = 30a. Equating this to 60 gives 30a=6030a = 60, so a=2a = 2. Therefore, the function is f(x)=2(x2)(x+3)(x5)f(x) = 2(x - 2)(x + 3)(x - 5). To find the value of f(1)f(1), substitute x=1x = 1 into this expression: f(1)=2(12)(1+3)(15)=2(1)(4)(4)=32f(1) = 2(1 - 2)(1 + 3)(1 - 5) = 2(-1)(4)(-4) = 32.

Adım Adım Çözüm

1
Identify the relation between the yy-intercept and the function's value.
f(0)=60f(0) = 60
The yy-intercept of a graph y=f(x)y = f(x) is the point where x=0x = 0.
2
Substitute x=0x = 0 into the definition of f(x)f(x) and set it equal to 60.
a(02)(0+3)(05)=60    30a=60a(0 - 2)(0 + 3)(0 - 5) = 60 \implies 30a = 60
This allows us to solve for the unknown constant coefficient aa.
3
Solve the linear equation for aa.
a=2a = 2
Dividing both sides of the equation by 30 isolates aa.
4
Evaluate the complete function f(x)=2(x2)(x+3)(x5)f(x) = 2(x - 2)(x + 3)(x - 5) at x=1x = 1.
f(1)=2(12)(1+3)(15)=2(1)(4)(4)=32f(1) = 2(1 - 2)(1 + 3)(1 - 5) = 2(-1)(4)(-4) = 32
This yields the requested value of f(1)f(1).

Anahtar Kavram

Using the factors and a known point (such as the y-intercept) of a polynomial function to determine its algebraic expression and evaluate it.
Soru 392Soru

The tables below show some values of the function ff and the transformed function gg, where g(x)=f(xh)+kg(x) = f(x - h) + k for constants hh and kk.

xxf(x)f(x)
2-288
0033
221-1
4455
xxg(x)g(x)
1166
3311
553-3
7733

What is the value of h+kh + k?

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Cevap: 1

Cevap

1
The correct answer is 1. By comparing the tables, each input of g(x)g(x) is 33 units greater than the corresponding input of f(x)f(x) (for example, g(1)g(1) corresponds to f(2)f(-2) because 1(2)=31 - (-2) = 3). A horizontal shift of 33 units to the right is represented in function notation by subtracting 33 from the input variable, so g(x)=f(x3)+kg(x) = f(x - 3) + k, which gives h=3h = 3. Next, comparing the output values shows that each output of g(x)g(x) is 22 units less than the corresponding output of f(x)f(x) (for example, g(1)=6g(1) = 6 while f(2)=8f(-2) = 8, and 6=826 = 8 - 2). This indicates a vertical shift downwards by 22 units, so k=2k = -2. Therefore, the value of h+kh + k is 3+(2)=13 + (-2) = 1.

Adım Adım Çözüm

1
Determine the horizontal translation constant hh by comparing the inputs of f(x)f(x) and g(x)g(x) that correspond to related output values.
The inputs for f(x)f(x) are {2,0,2,4}\{-2, 0, 2, 4\} and the corresponding inputs for g(x)g(x) are {1,3,5,7}\{1, 3, 5, 7\}. Each input for g(x)g(x) is 33 units greater than the corresponding input for f(x)f(x) (since 1(2)=31 - (-2) = 3, 30=33 - 0 = 3, etc.). This indicates a horizontal shift of 33 units to the right, which means the argument of ff in g(x)g(x) must be x3x - 3. Comparing this with f(xh)f(x - h) gives h=3h = 3.
Identifying the horizontal shift determines the value of the parameter hh in the translation equation g(x)=f(xh)+kg(x) = f(x - h) + k.
2
Determine the vertical translation constant kk by comparing the output values of g(x)g(x) with the corresponding values of f(x3)f(x - 3).
Using the matching inputs, compare the outputs:
- For x=1x = 1, g(1)=6g(1) = 6 and f(13)=f(2)=8f(1 - 3) = f(-2) = 8. The difference is 68=26 - 8 = -2.
- For x=3x = 3, g(3)=1g(3) = 1 and f(33)=f(0)=3f(3 - 3) = f(0) = 3. The difference is 13=21 - 3 = -2.
- For x=5x = 5, g(5)=3g(5) = -3 and f(53)=f(2)=1f(5 - 3) = f(2) = -1. The difference is 3(1)=2-3 - (-1) = -2.
- For x=7x = 7, g(7)=3g(7) = 3 and f(73)=f(4)=5f(7 - 3) = f(4) = 5. The difference is 35=23 - 5 = -2.
Since each output of g(x)g(x) is 22 units less than the corresponding output of f(x3)f(x - 3), the vertical translation is k=2k = -2.
Comparing the corresponding outputs identifies the vertical shift constant kk in the translation equation g(x)=f(xh)+kg(x) = f(x - h) + k.
3
Calculate the sum of the constants hh and kk.
h+k=3+(2)=1h + k = 3 + (-2) = 1.
Evaluating the sum of the parameters provides the final value requested by the question.

Anahtar Kavram

Function Notation and Transformations

Alternatif Yöntem

Instead of analyzing the general shift for all points, you can choose a single corresponding pair of points from the tables to set up equations. Let the point (2,8)(-2, 8) from the table of ff correspond to the point (1,6)(1, 6) from the table of gg. Since g(x)=f(xh)+kg(x) = f(x - h) + k, we substitute x=1x = 1 to get g(1)=f(1h)+kg(1) = f(1 - h) + k. Substituting the values g(1)=6g(1) = 6 and setting the input of ff to match, we get 1h=2    h=31 - h = -2 \implies h = 3. This simplifies the equation to 6=f(2)+k    6=8+k    k=26 = f(-2) + k \implies 6 = 8 + k \implies k = -2. This gives h+k=32=1h + k = 3 - 2 = 1.
Tahmini Süre:1m 30s
Soru 393Soru

The trajectory of a basketball thrown toward a hoop can be modeled by a quadratic function. In the xyxy-plane, xx represents the horizontal distance in feet from the shooter and yy represents the height of the basketball in feet. The basketball reaches its maximum height of 1414 feet at a horizontal distance of 88 feet from the shooter. If the basketball is released at a height of 66 feet, which of the following equations models the trajectory of the basketball?

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Cevap: y=18(x8)2+14y = -\frac{1}{8}(x - 8)^2 + 14

Cevap

The equation y=18(x8)2+14y = -\frac{1}{8}(x - 8)^2 + 14
The vertex form of a quadratic function is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) represents the coordinates of the vertex. Since the maximum height of the basketball is 1414 feet at a horizontal distance of 88 feet, the vertex is (8,14)(8, 14). Substituting these values gives the equation y=a(x8)2+14y = a(x - 8)^2 + 14. The release point represents the y-intercept where x=0x = 0 and y=6y = 6. Substituting these coordinates into the equation gives 6=a(08)2+146 = a(0 - 8)^2 + 14, which simplifies to 6=64a+146 = 64a + 14. Subtracting 1414 from both sides results in 8=64a-8 = 64a, and dividing both sides by 6464 yields a=18a = -\frac{1}{8}. Therefore, the correct trajectory model is y=18(x8)2+14y = -\frac{1}{8}(x - 8)^2 + 14.

Adım Adım Çözüm

1
Identify the vertex coordinates (h,k)(h, k) of the parabolic path from the given maximum height context.
The vertex (h,k)(h, k) is (8,14)(8, 14).
The basketball reaches its maximum height of 1414 feet at a horizontal distance of 88 feet, and the maximum of a downward-opening parabola is its vertex.
2
Write the general vertex form of a quadratic equation and substitute the coordinates of the vertex.
y=a(x8)2+14y = a(x - 8)^2 + 14
The vertex form of a quadratic equation is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
3
Substitute the initial release point (0,6)(0, 6) into the equation to solve for the coefficient aa.
6=a(08)2+14    6=64a+14    8=64a    a=864=186 = a(0 - 8)^2 + 14 \implies 6 = 64a + 14 \implies -8 = 64a \implies a = -\frac{8}{64} = -\frac{1}{8}
The basketball is released at a height of 66 feet when the horizontal distance x=0x = 0, representing the y-intercept of the trajectory.
4
Substitute the solved value of aa back into the vertex form equation.
y=18(x8)2+14y = -\frac{1}{8}(x - 8)^2 + 14
This yields the complete quadratic model representing the path of the basketball.

Anahtar Kavram

Writing quadratic equations in vertex form from context
Soru 394Soru

In the xyxy-plane, a cubic polynomial function ff crosses the xx-axis at (3,0)(-3, 0), (1,0)(1, 0), and (4,0)(4, 0). Given that f(2)=10f(2) = -10, what is the yy-coordinate of the yy-intercept of the graph of ff?

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Cevap: 12

Cevap

12
The correct answer is 12. Since the cubic function has xx-intercepts at (3,0)(-3, 0), (1,0)(1, 0), and (4,0)(4, 0), it can be represented in factored form as f(x)=a(x+3)(x1)(x4)f(x) = a(x + 3)(x - 1)(x - 4) for some constant aa. Substituting the given coordinate (2,10)(2, -10) into the function yields 10=a(2+3)(21)(24)-10 = a(2 + 3)(2 - 1)(2 - 4), which simplifies to 10=10a-10 = -10a, meaning a=1a = 1. The fully determined function is f(x)=(x+3)(x1)(x4)f(x) = (x + 3)(x - 1)(x - 4). The yy-intercept occurs when x=0x = 0. Evaluating f(0)f(0) gives (3)(1)(4)=12(3)(-1)(-4) = 12.

Adım Adım Çözüm

1
Write the general factored form of the cubic function using its intercepts.
f(x)=a(x+3)(x1)(x4)f(x) = a(x + 3)(x - 1)(x - 4), where aa is a constant.
Since the graph of ff has xx-intercepts at (3,0)(-3, 0), (1,0)(1, 0), and (4,0)(4, 0), the values 3-3, 11, and 44 are roots of the polynomial. Therefore, (x+3)(x + 3), (x1)(x - 1), and (x4)(x - 4) are factors of the polynomial.
2
Substitute the point (2,10)(2, -10) into the equation to solve for the constant coefficient aa.
a=1a = 1
Plugging in the given values yields 10=a(2+3)(21)(24)-10 = a(2 + 3)(2 - 1)(2 - 4), which simplifies to 10=a(5)(1)(2)-10 = a(5)(1)(-2), or 10=10a-10 = -10a. Dividing both sides by 10-10 gives a=1a = 1.
3
Calculate the yy-intercept of the function by evaluating it at x=0x = 0.
f(0)=12f(0) = 12
The yy-intercept occurs where the graph crosses the yy-axis, which corresponds to x=0x = 0. Substituting x=0x = 0 into the function f(x)=(x+3)(x1)(x4)f(x) = (x + 3)(x - 1)(x - 4) gives f(0)=(3)(1)(4)=12f(0) = (3)(-1)(-4) = 12.

Anahtar Kavram

Polynomial Factors and Graphs
Tahmini Süre:1m 30s
Soru 395Soru

The quadratic equation x2+10x+c=7x^2 + 10x + c = 7, where cc is a constant, has exactly one real solution. What is the value of cc?

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Cevap: 3232

Cevap

32
For the quadratic equation to have exactly one real solution, it must be written in the standard form x2+10x+(c7)=0x^2 + 10x + (c - 7) = 0, and its discriminant must equal zero. Setting the discriminant to zero gives 1024(1)(c7)=010^2 - 4(1)(c - 7) = 0. Distributing the negative four yields 1004c+28=0100 - 4c + 28 = 0, which simplifies to 1284c=0128 - 4c = 0. Solving for the constant gives c=32c = 32. Alternatively, the left side of the equation in standard form must be a perfect square trinomial, meaning x2+10x+25=0x^2 + 10x + 25 = 0. Thus, c7=25c - 7 = 25, which yields c=32c = 32.

Adım Adım Çözüm

1
Subtract 77 from both sides of the equation to write it in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2+10x+(c7)=0x^2 + 10x + (c - 7) = 0
To analyze the solutions of a quadratic equation using the discriminant, the equation must be in standard form.
2
Identify the coefficients and set the discriminant b24acb^2 - 4ac to 00 since the equation has exactly one real solution.
1024(1)(c7)=010^2 - 4(1)(c - 7) = 0
A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.
3
Solve the equation 1004(c7)=0100 - 4(c - 7) = 0 for cc by distributing the 4-4 and combining like terms.
1004c+28=0    1284c=0    4c=128    c=32100 - 4c + 28 = 0 \implies 128 - 4c = 0 \implies 4c = 128 \implies c = 32
Isolating the variable cc yields the value that makes the equation have exactly one real solution.

Anahtar Kavram

Discriminant of a quadratic equation

Alternatif Yöntem

Alternatively, you can complete the square. For the quadratic expression x2+10x+(c7)x^2 + 10x + (c - 7) to have exactly one real solution, it must be a perfect square trinomial of the form (x+d)2=x2+2dx+d2(x + d)^2 = x^2 + 2dx + d^2. Comparing coefficients, 2d=10    d=52d = 10 \implies d = 5, so the constant term must be d2=25d^2 = 25. Setting the constant term c7=25c - 7 = 25 yields c=32c = 32.
Tahmini Süre:1m 30s
Soru 396Soru

The solutions to the quadratic equation x26x11=0x^2 - 6x - 11 = 0 can be written in the form x=a±bx = a \pm \sqrt{b}, where aa and bb are integers. What is the value of a+ba + b?

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Cevap: 23

Cevap

23
Completing the square on the quadratic equation x26x11=0x^2 - 6x - 11 = 0 yields the solutions x=3±20x = 3 \pm \sqrt{20}. Comparing this to the given form x=a±bx = a \pm \sqrt{b} identifies a=3a = 3 and b=20b = 20. Summing these values gives a+b=23a + b = 23.

Adım Adım Çözüm

1
Add 11 to both sides of the equation to isolate the variable terms.
x26x=11x^2 - 6x = 11
Preparing the quadratic equation to complete the square by separating constant terms.
2
Complete the square by adding the square of half the coefficient of xx to both sides.
x26x+9=11+9x^2 - 6x + 9 = 11 + 9, which simplifies to (x3)2=20(x - 3)^2 = 20
Adding (62)2=9(\frac{-6}{2})^2 = 9 creates a perfect square trinomial on the left side.
3
Take the square root of both sides and solve for xx.
x3=±20x - 3 = \pm\sqrt{20}, which gives x=3±20x = 3 \pm \sqrt{20}
Taking the square root cancels the exponent and yields two possible solutions.
4
Compare the solutions to the given form x=a±bx = a \pm \sqrt{b} to determine aa and bb, and calculate their sum.
a=3a = 3, b=20b = 20, so a+b=3+20=23a + b = 3 + 20 = 23
Matching the template shows that aa corresponds to 33 and bb corresponds to 2020.

Anahtar Kavram

Solving quadratic equations by completing the square and matching solutions to a given form.

Alternatif Yöntem

Instead of completing the square, you can use the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=1a = 1, b=6b = -6, and c=11c = -11. This yields x=6±364(1)(11)2=6±802=3±802=3±20x = \frac{6 \pm \sqrt{36 - 4(1)(-11)}}{2} = \frac{6 \pm \sqrt{80}}{2} = 3 \pm \frac{\sqrt{80}}{2} = 3 \pm \sqrt{20}. Comparing this to the given form yields a=3a = 3 and b=20b = 20, which sums to 2323.
Tahmini Süre:1m 30s
Soru 397Soru

In the quadratic equation 2x215x+c=02x^2 - 15x + c = 0, cc is a constant. If one of the solutions to the equation is x=6x = 6, what is the other solution?

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Cevap: 1.5

Cevap

1.5
The sum of the roots of a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0 is given by ba-\frac{b}{a}. For the equation 2x215x+c=02x^2 - 15x + c = 0, the sum of the roots is 152=7.5-\frac{-15}{2} = 7.5. Since one of the solutions is 66, the other solution is 7.56=1.57.5 - 6 = 1.5. Alternatively, substituting x=6x = 6 into the equation yields 2(6)215(6)+c=0    7290+c=0    c=182(6)^2 - 15(6) + c = 0 \implies 72 - 90 + c = 0 \implies c = 18. The equation becomes 2x215x+18=02x^2 - 15x + 18 = 0, which factors as (2x3)(x6)=0(2x - 3)(x - 6) = 0. Setting the factor 2x3=02x - 3 = 0 gives the other solution, x=1.5x = 1.5 (or 32\frac{3}{2}).

Adım Adım Çözüm

1
Find the sum of the roots of the quadratic equation using Vieta's formulas.
The sum of the roots is 152=7.5-\frac{-15}{2} = 7.5.
For any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of its roots is equal to ba-\frac{b}{a}.
2
Set up an equation using the known root to find the unknown root.
6+x2=7.56 + x_2 = 7.5, where x2x_2 is the other root.
Since one root is 66, adding it to the second root must equal the sum of the roots.
3
Solve for the second root.
x2=1.5x_2 = 1.5.
Subtracting 66 from both sides of the equation yields 1.51.5.

Anahtar Kavram

Sum of roots of a quadratic equation (Vieta's Formulas)
Tahmini Süre:1m 30s
Soru 398Soru

The function ff is defined by f(x)=x2+6x1f(x) = -x^2 + 6x - 1. The function gg is defined by g(x)=f(x+2)5g(x) = f(x + 2) - 5. If the maximum value of g(x)g(x) in the xyxy-plane occurs at the point (h,k)(h, k), what is the value of h+kh + k?

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Cevap: 4

Cevap

4
The vertex of the original quadratic function f(x)=x2+6x1f(x) = -x^2 + 6x - 1 is located at (3,8)(3, 8). The transformation g(x)=f(x+2)5g(x) = f(x + 2) - 5 translates the graph horizontally to the left by 22 units and vertically down by 55 units. This moves the vertex from (3,8)(3, 8) to (32,85)=(1,3)(3 - 2, 8 - 5) = (1, 3). Therefore, h=1h = 1 and k=3k = 3, and their sum h+kh + k equals 44.

Adım Adım Çözüm

1
Find the vertex of the function f(x)=x2+6x1f(x) = -x^2 + 6x - 1.
The vertex of f(x)f(x) is at (3,8)(3, 8).
By writing f(x)f(x) in vertex form, f(x)=(x3)2+8f(x) = -(x - 3)^2 + 8, we find that the maximum value of f(x)f(x) occurs at (3,8)(3, 8).
2
Determine the vertex (h,k)(h, k) of the transformed function g(x)=f(x+2)5g(x) = f(x + 2) - 5.
(h,k)=(1,3)(h, k) = (1, 3)
The horizontal shift of f(x+2)f(x + 2) translates the graph to the left by 22 units, changing the xx-coordinate from 33 to 32=13 - 2 = 1. The vertical shift of 5- 5 translates the graph down by 55 units, changing the yy-coordinate from 88 to 85=38 - 5 = 3.
3
Calculate the value of h+kh + k.
4
Adding the coordinates of the transformed vertex yields h+k=1+3=4h + k = 1 + 3 = 4.

Anahtar Kavram

Vertex form of a quadratic function and translation of functions.
Soru 399Soru

In the quadratic equation 3x218x+c=03x^2 - 18x + c = 0, cc is a constant. If the sum of the squares of the two real solutions to this equation is 2626, what is the value of cc?

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Cevap: 15

Cevap

The value of cc is 1515.
To find the value of cc, we apply Vieta's formulas to the equation 3x218x+c=03x^2 - 18x + c = 0. The sum of the solutions is x1+x2=183=6x_1 + x_2 = -\frac{-18}{3} = 6, and the product of the solutions is x1x2=c3x_1 x_2 = \frac{c}{3}. Using the identity x12+x22=(x1+x2)22x1x2x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2, we substitute the given values: 26=622(c3)26 = 6^2 - 2\left(\frac{c}{3}\right). This simplifies to 26=362c326 = 36 - \frac{2c}{3}. Subtracting 3636 from both sides yields 10=2c3-10 = -\frac{2c}{3}, which gives 2c=302c = 30, and thus c=15c = 15. Alternatively, since the vertex of the corresponding parabola is at x=182(3)=3x = -\frac{-18}{2(3)} = 3, the two solutions can be represented symmetrically as 3+d3 + d and 3d3 - d. The sum of their squares is (3+d)2+(3d)2=(9+6d+d2)+(96d+d2)=18+2d2(3 + d)^2 + (3 - d)^2 = (9 + 6d + d^2) + (9 - 6d + d^2) = 18 + 2d^2. Setting this equal to the given value of 2626 yields 18+2d2=26    2d2=8    d2=4    d=218 + 2d^2 = 26 \implies 2d^2 = 8 \implies d^2 = 4 \implies d = 2. Therefore, the solutions are 3+2=53 + 2 = 5 and 32=13 - 2 = 1. Substituting either solution back into the original equation, such as x=1x = 1, yields 3(1)218(1)+c=0    15+c=0    c=153(1)^2 - 18(1) + c = 0 \implies -15 + c = 0 \implies c = 15.

Adım Adım Çözüm

1
Find the sum and product of the solutions to the quadratic equation.
The sum of the solutions is x1+x2=6x_1 + x_2 = 6, and their product is x1x2=c3x_1 x_2 = \frac{c}{3}.
According to Vieta's formulas, for a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the solutions is ba-\frac{b}{a} and the product of the solutions is ca\frac{c}{a}.
2
Relate the sum of the squares of the solutions to their sum and product using the algebraic identity.
x12+x22=(x1+x2)22x1x2    26=(6)22(c3)x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 \implies 26 = (6)^2 - 2\left(\frac{c}{3}\right).
The algebraic identity (x1+x2)2=x12+2x1x2+x22(x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2 can be rearranged to express the sum of squares x12+x22x_1^2 + x_2^2 in terms of the sum and product.
3
Solve the resulting linear equation for the constant cc.
26=362c3    2c3=10    c=1526 = 36 - \frac{2c}{3} \implies \frac{2c}{3} = 10 \implies c = 15.
Subtracting 3636 from both sides gives 10=2c3-10 = -\frac{2c}{3}. Multiplying by 3-3 gives 30=2c30 = 2c, which yields c=15c = 15.

Anahtar Kavram

Vieta's Formulas and Algebraic Identities
Soru 400Soru

In the xyxy-plane, the graph of the quadratic function ff is a parabola with vertex (3,8)(3, -8) and passes through the point (1,0)(1, 0). If the graph intersects the yy-axis at (0,c)(0, c), what is the value of cc?

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Cevap: 10

Cevap

The correct answer is 10.
By writing the function in vertex form f(x)=a(xh)2+kf(x) = a(x - h)^2 + k and substituting the vertex (3,8)(3, -8), we obtain f(x)=a(x3)28f(x) = a(x - 3)^2 - 8. Substituting the point (1,0)(1, 0) into the equation gives 0=a(13)280 = a(1 - 3)^2 - 8, which simplifies to 4a=84a = 8 and yields a=2a = 2. With the function fully defined as f(x)=2(x3)28f(x) = 2(x - 3)^2 - 8, the yy-intercept is found by calculating f(0)=2(03)28=2(9)8=10f(0) = 2(0 - 3)^2 - 8 = 2(9) - 8 = 10. Thus, c=10c = 10.

Adım Adım Çözüm

1
Express the quadratic function in vertex form using the given vertex (3,8)(3, -8).
f(x)=a(x3)28f(x) = a(x - 3)^2 - 8
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 point (1,0)(1, 0) into the function to solve for the leading coefficient aa.
a=2a = 2
Since the graph passes through (1,0)(1, 0), substituting x=1x = 1 and f(x)=0f(x) = 0 must satisfy the equation.
3
Substitute x=0x = 0 into the completed function f(x)=2(x3)28f(x) = 2(x - 3)^2 - 8 to find the value of cc.
c=10c = 10
The intersection with the yy-axis occurs where the input xx is 0.

Anahtar Kavram

Writing and evaluating quadratic functions using the vertex form
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Advanced Math Alıştırma Soruları — SAT — Sayfa 20 | Examkin