Quadratic Functions and Graphs

76 soru

Soru 41Soru

The graph of the quadratic function f(x)=2(x3)2+af(x) = -2(x - 3)^2 + a in the xyxy-plane has a yy-intercept at (0,10)(0, -10), where aa is a constant. What is the maximum value of f(x)f(x)?

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Cevap: 88

Cevap

The maximum value of the function is 88.
The correct answer is 88. Since the yy-intercept of the graph is (0,10)(0, -10), we substitute x=0x = 0 into the function: f(0)=2(03)2+a=10f(0) = -2(0 - 3)^2 + a = -10. Simplifying this yields 2(9)+a=10-2(9) + a = -10, which becomes 18+a=10-18 + a = -10. Adding 1818 to both sides gives a=8a = 8. The equation of the function is therefore f(x)=2(x3)2+8f(x) = -2(x - 3)^2 + 8. Since this is in vertex form, the vertex is (3,8)(3, 8). Because the coefficient of the squared term is negative, the parabola opens downward, and the maximum value of the function is the yy-coordinate of the vertex, which is 88.

Adım Adım Çözüm

1
Use the given yy-intercept to set up an equation for the constant aa.
f(0)=10    2(03)2+a=10f(0) = -10 \implies -2(0 - 3)^2 + a = -10
The yy-intercept occurs where the input xx is equal to 00.
2
Simplify the expression and solve for aa.
2(3)2+a=10    2(9)+a=10    18+a=10    a=8-2(-3)^2 + a = -10 \implies -2(9) + a = -10 \implies -18 + a = -10 \implies a = 8
Squaring 3-3 yields 99, and multiplying by 2-2 gives 18-18. Adding 1818 to both sides isolates aa.
3
Identify the vertex of the quadratic function and determine the maximum value.
The function is f(x)=2(x3)2+8f(x) = -2(x - 3)^2 + 8. The vertex of this parabola is (3,8)(3, 8). Since the leading coefficient 2-2 is negative, the parabola opens downward, meaning the yy-coordinate of the vertex, 88, is the maximum value of f(x)f(x).
A quadratic function in vertex form f(x)=a(xh)2+kf(x) = a(x - h)^2 + k has its vertex at (h,k)(h, k). If a<0a < 0, the maximum value is kk.

Anahtar Kavram

Identifying the vertex and maximum value of a quadratic function from its vertex form and yy-intercept.
Soru 42Soru

A quadratic function ff has a vertex at (h,k)(h, k), where hh and kk are constants. In the xyxy-plane, the graph of y=f(x)y = f(x) contains the points (1,3)(-1, 3) and (7,3)(7, 3). The function gg is defined by g(x)=f(x+2)4g(x) = f(x + 2) - 4, and its graph has a vertex at (p,q)(p, q). If p+q=5p + q = 5, what is the value of kk?

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Cevap: 8

Cevap

The value of k is 8.
The axis of symmetry of the quadratic function f is located at the midpoint of the x-values of any two points on the graph that have the same y-value. Since the graph of f contains (-1, 3) and (7, 3), the x-coordinate of its vertex, h, is equal to the midpoint (-1 + 7) / 2 = 3. Thus, the vertex of f is (3, k). The definition of g(x) = f(x + 2) - 4 indicates a horizontal translation of 2 units to the left and a vertical translation of 4 units down. Applying these translations to the vertex (3, k) of f gives the vertex of g as (3 - 2, k - 4) = (1, k - 4). Since the vertex of g is defined as (p, q), we have p = 1 and q = k - 4. Substituting these values into the equation p + q = 5 yields 1 + (k - 4) = 5, which simplifies to k - 3 = 5, or k = 8.

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1
Find the x-coordinate of the vertex of the function f
h = 3
Because the points (-1, 3) and (7, 3) have the same y-coordinate, the axis of symmetry (and thus the x-coordinate of the vertex) must lie exactly halfway between x = -1 and x = 7.
2
Determine the vertex of the function g(x)
(p, q) = (1, k - 4)
The function g(x) = f(x + 2) - 4 represents a translation of the graph of f by 2 units to the left and 4 units down. Since the vertex of f is (3, k), the vertex of g is (3 - 2, k - 4).
3
Solve for k using the given sum p + q = 5
k = 8
Substitute p = 1 and q = k - 4 into the equation p + q = 5 to get 1 + k - 4 = 5, which simplifies to k = 8.

Anahtar Kavram

Quadratic functions and graphs, including symmetry and translations.
Soru 43Soru

In the xyxy-plane, the graph of the quadratic function f(x)=2(xd)2+8f(x) = -2(x - d)^2 + 8, where dd is a positive constant, intersects the xx-axis at the point (2,0)(2, 0). The function gg is defined by g(x)=f(x+3)5g(x) = f(x + 3) - 5. What is the yy-coordinate of the yy-intercept of the graph of gg in the xyxy-plane?

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Cevap: 1

Cevap

The yy-coordinate of the yy-intercept of the graph of gg is 11.
To find the correct answer, we first substitute the point (2,0)(2,0) into f(x)=2(xd)2+8f(x) = -2(x - d)^2 + 8 to obtain 0=2(2d)2+80 = -2(2 - d)^2 + 8. Simplifying this equation gives (2d)2=4(2 - d)^2 = 4, which has solutions d=0d = 0 and d=4d = 4. Since the problem states that dd is a positive constant, we have d=4d = 4, which means f(x)=2(x4)2+8f(x) = -2(x - 4)^2 + 8. We then find the equation for the transformed function g(x)g(x) by replacing xx with x+3x + 3 and subtracting 55, which yields g(x)=f(x+3)5=2(x+34)2+85=2(x1)2+3g(x) = f(x + 3) - 5 = -2(x + 3 - 4)^2 + 8 - 5 = -2(x - 1)^2 + 3. To find the yy-coordinate of the yy-intercept, we evaluate g(x)g(x) at x=0x = 0, giving g(0)=2(01)2+3=2(1)+3=1g(0) = -2(0 - 1)^2 + 3 = -2(1) + 3 = 1.

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1
Use the given xx-intercept (2,0)(2,0) of f(x)f(x) to solve for the constant dd.
d=4d = 4
Substitute x=2x = 2 and f(x)=0f(x) = 0 into the equation f(x)=2(xd)2+8f(x) = -2(x - d)^2 + 8 to get 0=2(2d)2+80 = -2(2 - d)^2 + 8. Dividing by 2-2 and isolating the squared term gives (2d)2=4(2 - d)^2 = 4. Taking the square root of both sides gives 2d=22 - d = 2 or 2d=22 - d = -2, which yields d=0d = 0 or d=4d = 4. Since dd must be a positive constant, we must choose d=4d = 4.
2
Determine the equation of the transformed function g(x)g(x).
g(x)=2(x1)2+3g(x) = -2(x - 1)^2 + 3
Substitute d=4d = 4 into f(x)f(x) to get f(x)=2(x4)2+8f(x) = -2(x - 4)^2 + 8. Since g(x)=f(x+3)5g(x) = f(x + 3) - 5, substitute x+3x + 3 for xx in f(x)f(x) and subtract 55: g(x)=2((x+3)4)2+85g(x) = -2((x + 3) - 4)^2 + 8 - 5, which simplifies to g(x)=2(x1)2+3g(x) = -2(x - 1)^2 + 3.
3
Find the yy-coordinate of the yy-intercept of the graph of gg.
11
The yy-intercept of a graph is the point where x=0x = 0. Substitute x=0x = 0 into the equation for g(x)g(x) to find g(0)=2(01)2+3=2(1)+3=1g(0) = -2(0 - 1)^2 + 3 = -2(1) + 3 = 1.

Anahtar Kavram

Applying horizontal and vertical translations to quadratic equations and identifying graph features.
Soru 44Soru

A quadratic function ff is defined by f(x)=a(x4)(x10)f(x) = a(x - 4)(x - 10), where aa is a positive constant. In the xyxy-plane, the graph of ff has a vertex with a yy-coordinate of 18-18. What is the value of aa?

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Cevap: 2

Cevap

The value of aa is 2.
The vertex of a parabola lies on the axis of symmetry, which is located midway between its x-intercepts. Since the function is f(x)=a(x4)(x10)f(x) = a(x - 4)(x - 10), the x-intercepts are at x=4x = 4 and x=10x = 10. The midpoint of these values is 4+102=7\frac{4 + 10}{2} = 7. Thus, the x-coordinate of the vertex is 7. Since the y-coordinate of the vertex is given as 18-18, the vertex is at (7,18)(7, -18). Substituting these coordinates into the function equation yields 18=a(74)(710)-18 = a(7 - 4)(7 - 10), which simplifies to 18=a(3)(3)-18 = a(3)(-3), or 18=9a-18 = -9a. Solving for aa gives a=2a = 2.

Adım Adım Çözüm

1
Find the x-coordinate of the vertex using the symmetry of the parabola.
The x-coordinate of the vertex is 7.
The x-coordinate of the vertex lies on the axis of symmetry, which is the midpoint of the x-intercepts x=4x = 4 and x=10x = 10.
2
Identify the coordinates of the vertex.
The vertex is at (7,18)(7, -18).
The y-coordinate of the vertex is given as 18-18, and its x-coordinate was calculated as 7.
3
Substitute the vertex coordinates into the function definition to solve for aa.
a=2a = 2
Substituting (7,18)(7, -18) into f(x)=a(x4)(x10)f(x) = a(x - 4)(x - 10) gives 18=a(74)(710)-18 = a(7 - 4)(7 - 10), which simplifies to 18=9a-18 = -9a.

Anahtar Kavram

Finding the vertex of a quadratic function from its factored form and solving for a leading coefficient.

Alternatif Yöntem

Alternatively, the function f(x)=a(x4)(x10)f(x) = a(x - 4)(x - 10) can be expanded to standard form: f(x)=a(x214x+40)f(x) = a(x^2 - 14x + 40). The x-coordinate of the vertex can be found using the formula x=b2ax = -\frac{b}{2a}, which gives x=14a2a=7x = -\frac{-14a}{2a} = 7. Then, substitute x=7x = 7 and f(7)=18f(7) = -18 to solve for aa.
Tahmini Süre:1m 30s
Soru 45Soru

A quadratic function gg is defined by g(x)=2x212x+kg(x) = 2x^2 - 12x + k, where kk is a constant. In the xyxy-plane, the graph of gg has its vertex on the line y=5y = -5. What is the value of kk?

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Cevap: 13

Cevap

13
The xx-coordinate of the vertex for the quadratic function g(x)=ax2+bx+cg(x) = ax^2 + bx + c is found using the formula x=b2ax = -\frac{b}{2a}. Substituting a=2a = 2 and b=12b = -12 yields x=122(2)=3x = -\frac{-12}{2(2)} = 3. Evaluating g(3)g(3) gives the yy-coordinate of the vertex: g(3)=2(3)212(3)+k=18+kg(3) = 2(3)^2 - 12(3) + k = -18 + k. Since the vertex lies on the line y=5y = -5, setting 18+k=5-18 + k = -5 and solving for kk gives k=13k = 13.

Adım Adım Çözüm

1
Find the xx-coordinate of the vertex of the parabola.
The xx-coordinate is 33.
For a quadratic function g(x)=ax2+bx+cg(x) = ax^2 + bx + c, the xx-coordinate of the vertex is given by x=b2ax = -\frac{b}{2a}. Here, a=2a = 2 and b=12b = -12, so x=122(2)=3x = -\frac{-12}{2(2)} = 3.
2
Evaluate the function at x=3x = 3 to find the yy-coordinate of the vertex in terms of kk.
The yy-coordinate is 18+k-18 + k.
Substituting x=3x = 3 into the function gives g(3)=2(3)212(3)+k=1836+k=18+kg(3) = 2(3)^2 - 12(3) + k = 18 - 36 + k = -18 + k.
3
Set the yy-coordinate of the vertex equal to 5-5 and solve for kk.
k=13k = 13.
Since the vertex lies on the line y=5y = -5, the yy-coordinate of the vertex must be 5-5. Thus, 18+k=5    k=13-18 + k = -5 \implies k = 13.

Anahtar Kavram

Determining the vertex of a quadratic function and using its coordinates to solve for unknown parameters.
Soru 46Soru

The graph of the quadratic function ff in the xyxy-plane has its vertex at (3,4)(3, -4) and passes through the point (1,8)(1, 8). The function gg is defined by g(x)=f(xh)+kg(x) = f(x - h) + k, where hh and kk are constants. If the graph of gg has its vertex at (0,0)(0, 0), what is the value of g(4)g(4)?

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Cevap: 48

Cevap

48
The quadratic function f(x)f(x) can be written in vertex form as f(x)=a(x3)24f(x) = a(x - 3)^2 - 4. Substituting the point (1,8)(1, 8) gives 8=a(13)248 = a(1 - 3)^2 - 4, which simplifies to 12=4a12 = 4a, so a=3a = 3. This means f(x)=3(x3)24f(x) = 3(x - 3)^2 - 4. The function g(x)=f(xh)+kg(x) = f(x - h) + k has its vertex at (0,0)(0, 0), which is a translation of the vertex of f(x)f(x) from (3,4)(3, -4) to (0,0)(0, 0). This shifts the function to g(x)=3x2g(x) = 3x^2. Evaluating g(4)g(4) yields 3(4)2=3(16)=483(4)^2 = 3(16) = 48.

Adım Adım Çözüm

1
Write the vertex form of the quadratic function f(x)f(x).
f(x)=a(x3)24f(x) = a(x - 3)^2 - 4
The vertex of the parabola is given as (3,4)(3, -4), so we can substitute these coordinates into the vertex form f(x)=a(xh)2+kf(x) = a(x - h)^2 + k.
2
Substitute the point (1,8)(1, 8) into the equation to solve for the constant aa.
8=a(13)24    12=4a    a=38 = a(1 - 3)^2 - 4 \implies 12 = 4a \implies a = 3
Since the graph passes through (1,8)(1, 8), these coordinates must satisfy the equation of the function.
3
Determine the equation of the transformed function g(x)g(x).
g(x)=3x2g(x) = 3x^2
The vertex of f(x)f(x) is at (3,4)(3, -4). Shifting this vertex to (0,0)(0, 0) removes the horizontal and vertical offsets, simplifying the function to g(x)=ax2g(x) = ax^2 with the same leading coefficient a=3a = 3.
4
Evaluate g(4)g(4).
g(4)=3(4)2=3(16)=48g(4) = 3(4)^2 = 3(16) = 48
Substitute x=4x = 4 into the equation for g(x)g(x) to find the required value.

Anahtar Kavram

Quadratic Functions and Graphs
Soru 47Soru

A parabola in the xyxy-plane has vertex (3,18)(3, 18) and passes through the origin. If the equation of the parabola is written in the form y=ax2+bx+cy = ax^2 + bx + c, where aa, bb, and cc are constants, what is the value of a+ba + b?

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Cevap: 10

Cevap

10
The vertex form of a quadratic function with vertex (h,k)(h, k) is y=a(xh)2+ky = a(x - h)^2 + k. Substituting the given vertex (3,18)(3, 18) yields the equation y=a(x3)2+18y = a(x - 3)^2 + 18. Since the parabola passes through the origin, we can substitute the point (0,0)(0, 0) into the equation to find the value of aa: 0=a(03)2+18    9a=18    a=20 = a(0 - 3)^2 + 18 \implies 9a = -18 \implies a = -2. Substituting a=2a = -2 back into the vertex form and expanding gives y=2(x3)2+18=2(x26x+9)+18=2x2+12xy = -2(x - 3)^2 + 18 = -2(x^2 - 6x + 9) + 18 = -2x^2 + 12x. Comparing this to the standard form y=ax2+bx+cy = ax^2 + bx + c, we identify a=2a = -2 and b=12b = 12. The sum of these constants is a+b=2+12=10a + b = -2 + 12 = 10.

Adım Adım Çözüm

1
Write the equation of the parabola in vertex form.
y=a(x3)2+18y = a(x - 3)^2 + 18
The vertex form of a quadratic function is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
2
Substitute the coordinates of the origin (0,0)(0, 0) to solve for aa.
a=2a = -2
The parabola passes through the origin (0,0)(0, 0), so these coordinates must satisfy the equation.
3
Expand the vertex form equation into standard form y=ax2+bx+cy = ax^2 + bx + c.
y=2x2+12xy = -2x^2 + 12x
Expanding the equation allows us to identify the coefficients aa and bb directly.
4
Identify the values of aa and bb, and calculate a+ba + b.
a+b=10a + b = 10
Comparing y=2x2+12xy = -2x^2 + 12x to y=ax2+bx+cy = ax^2 + bx + c gives a=2a = -2 and b=12b = 12, so a+b=2+12=10a + b = -2 + 12 = 10.

Anahtar Kavram

Vertex form of quadratic functions and conversion to standard form
Soru 48Soru

In the xyxy-plane, the graph of the quadratic function ff has its vertex at the point (4,3)(4, -3). The function gg is defined by g(x)=f(x+2)+5g(x) = f(x + 2) + 5. Which of the following ordered pairs represents the vertex of the graph of gg?

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Cevap: (2,2)(2, 2)

Cevap

(2,2)(2, 2)
The vertex of f(x)f(x) is given as (4,3)(4, -3). The translation g(x)=f(x+2)+5g(x) = f(x + 2) + 5 shifts the graph 22 units to the left and 55 units upward. Applying these transformations to the vertex of f(x)f(x) yields the new vertex (42,3+5)=(2,2)(4 - 2, -3 + 5) = (2, 2).

Adım Adım Çözüm

1
Identify the vertex of the function f(x)f(x)
The vertex of f(x)f(x) is (4,3)(4, -3)
This is given in the problem and serves as the starting point for the transformations.
2
Determine the effect of the horizontal translation f(x+2)f(x + 2)
The vertex shifts 22 units to the left, changing the xx-coordinate from 44 to 42=24 - 2 = 2
For any function f(x)f(x), the graph of f(x+c)f(x + c) is shifted cc units to the left when c>0c > 0.
3
Determine the effect of the vertical translation +5+ 5
The vertex shifts 55 units upward, changing the yy-coordinate from 3-3 to 3+5=2-3 + 5 = 2
Adding a constant to a function shifts its graph vertically upward by that constant's value.
4
Combine the translated coordinates to find the vertex of g(x)g(x)
The vertex of the graph of g(x)g(x) is (2,2)(2, 2)
The horizontal shift results in an xx-coordinate of 22, and the vertical shift results in a yy-coordinate of 22.

Anahtar Kavram

Quadratic Transformations and Vertex Shifts
Soru 49Soru

The graph of the quadratic function ff in the xyxy-plane has xx-intercepts at (2,0)(-2, 0) and (8,0)(8, 0). If the maximum value of f(x)f(x) is 2525, what is the value of f(0)f(0)?

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Cevap: 16

Cevap

16
The axis of symmetry of the quadratic function ff lies halfway between the xx-intercepts x=2x = -2 and x=8x = 8, which is at x=2+82=3x = \frac{-2 + 8}{2} = 3. Since the function has a maximum value of 2525, this maximum must occur at the vertex, giving the vertex coordinates (3,25)(3, 25). In vertex form, the function is f(x)=a(x3)2+25f(x) = a(x - 3)^2 + 25. Substituting the xx-intercept (8,0)(8, 0) into the function yields 0=a(83)2+250 = a(8 - 3)^2 + 25, which simplifies to 25a=2525a = -25, or a=1a = -1. Therefore, the equation of the function is f(x)=(x3)2+25f(x) = -(x - 3)^2 + 25. Evaluating this at x=0x = 0 gives f(0)=(03)2+25=9+25=16f(0) = -(0 - 3)^2 + 25 = -9 + 25 = 16.

Adım Adım Çözüm

1
Find the xx-coordinate of the vertex (axis of symmetry)
x=3x = 3
The axis of symmetry of a parabola is located exactly halfway between its xx-intercepts: x=2+82=3x = \frac{-2 + 8}{2} = 3.
2
Determine the vertex coordinates
(3,25)(3, 25)
The maximum value of the quadratic function occurs at its vertex, so the yy-coordinate of the vertex is the maximum value 2525.
3
Write the vertex form of the quadratic function
f(x)=a(x3)2+25f(x) = a(x - 3)^2 + 25
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.
4
Solve for the leading coefficient aa
a=1a = -1
Substitute the xx-intercept (8,0)(8, 0) into the vertex form: 0=a(83)2+25    25a=25    a=10 = a(8 - 3)^2 + 25 \implies 25a = -25 \implies a = -1.
5
Find the value of f(0)f(0)
f(0)=16f(0) = 16
Substitute x=0x = 0 into the function: f(0)=(03)2+25=9+25=16f(0) = -(0 - 3)^2 + 25 = -9 + 25 = 16.

Anahtar Kavram

Using xx-intercepts and the maximum value to determine the vertex and equation of a quadratic function.
Soru 50Soru

An object is launched from a platform. The function h(t)=5t2+30t+12h(t) = -5t^2 + 30t + 12 models the height of the object, in meters, tt seconds after it was launched. What is the maximum height, in meters, reached by the object?

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Cevap: 57

Cevap

The maximum height reached by the object is 57 meters.
The given function h(t)=5t2+30t+12h(t) = -5t^2 + 30t + 12 represents a parabola opening downward because the coefficient of t2t^2 is negative. The maximum value of this function occurs at its vertex. The time tt at the vertex is determined using the formula t=b2a=302(5)=3t = -\frac{b}{2a} = -\frac{30}{2(-5)} = 3 seconds. Substituting t=3t = 3 back into the function yields the maximum height: h(3)=5(3)2+30(3)+12=45+90+12=57h(3) = -5(3)^2 + 30(3) + 12 = -45 + 90 + 12 = 57 meters.

Adım Adım Çözüm

1
Identify the coefficients of the quadratic function in standard form h(t)=at2+bt+ch(t) = at^2 + bt + c.
a=5a = -5, b=30b = 30, and c=12c = 12.
These coefficients are needed to calculate the vertex of the parabola.
2
Calculate the time tt at which the maximum height occurs using the vertex formula t=b2at = -\frac{b}{2a}.
t=302(5)=3t = -\frac{30}{2(-5)} = 3 seconds.
Since the leading coefficient a=5a = -5 is negative, the parabola opens downward, meaning its vertex represents the maximum value.
3
Substitute t=3t = 3 back into the height function to find the maximum height.
h(3)=5(3)2+30(3)+12=45+90+12=57h(3) = -5(3)^2 + 30(3) + 12 = -45 + 90 + 12 = 57 meters.
Evaluating the function at the time of the vertex gives the corresponding maximum height.

Anahtar Kavram

The maximum value of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c (where a<0a < 0) is the vertical coordinate of its vertex, which occurs at x=b2ax = -\frac{b}{2a}.
Soru 51Soru

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 (4,25)(4, 25). If the positive xx-intercept of the graph of ff is (d,0)(d, 0), what is the value of dd?

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Cevap: 9

Cevap

The value of dd is 99.
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. Since the vertex is (4,25)(4, 25) and the coefficient of x2x^2 is 1-1, the function is f(x)=(x4)2+25f(x) = -(x-4)^2 + 25. Setting f(x)=0f(x) = 0 to find the xx-intercepts yields (x4)2+25=0-(x-4)^2 + 25 = 0, which simplifies to (x4)2=25(x-4)^2 = 25. Taking the square root of both sides gives x4=5x - 4 = 5 or x4=5x - 4 = -5. Solving these equations gives x=9x = 9 or x=1x = -1. The positive xx-intercept is (9,0)(9, 0), so the value of dd is 99.

Adım Adım Çözüm

1
Write the function in vertex form.
f(x)=(x4)2+25f(x) = -(x-4)^2 + 25
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. Here, the vertex is (4,25)(4, 25) and the coefficient of x2x^2 is 1-1, so a=1a = -1, h=4h = 4, and k=25k = 25.
2
Set f(x)=0f(x) = 0 to find the xx-intercepts.
(x4)2+25=0-(x-4)^2 + 25 = 0
The xx-intercepts of a graph are the points where the function value is equal to 0.
3
Solve the equation for xx.
x=9x = 9 or x=1x = -1
Rearranging the equation gives (x4)2=25(x-4)^2 = 25. Taking the square root of both sides gives x4=5x-4 = 5 or x4=5x-4 = -5, which solves to x=9x = 9 or x=1x = -1.
4
Identify the positive xx-intercept coordinate dd.
d=9d = 9
The question asks for the positive xx-intercept (d,0)(d, 0), which corresponds to x=9x = 9.

Anahtar Kavram

Vertex form of a quadratic function and finding xx-intercepts
Soru 52Soru

The graph of the quadratic function ff in the xyxy-plane has its vertex at (2,5)(2, -5). If the graph passes through the point (5,13)(5, 13), what is the value of f(1)f(-1)?

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Cevap: 13

Cevap

13
The vertex of the parabola is (2,5)(2, -5), which indicates that the axis of symmetry is the vertical line x=2x = 2. The given point has an xx-coordinate of 55, which is a distance of 33 units from the axis of symmetry (52=35 - 2 = 3). The target point has an xx-coordinate of 1-1, which is also a distance of 33 units from the axis of symmetry (2(1)=32 - (-1) = 3). Because a parabola is perfectly symmetric about its axis of symmetry, any two points that are the same horizontal distance from this line must have the same yy-coordinate. Thus, f(1)f(-1) must be equal to f(5)f(5), which is 1313. Alternatively, one can find the specific equation of the quadratic function by substituting the vertex and the point (5,13)(5, 13) into the vertex form f(x)=a(x2)25f(x) = a(x - 2)^2 - 5, yielding a=2a = 2. Evaluating f(1)=2(12)25f(-1) = 2(-1 - 2)^2 - 5 gives 1313.

Adım Adım Çözüm

1
Identify the axis of symmetry from the given vertex.
The axis of symmetry is x=2x = 2.
For any quadratic function with a vertex at (h,k)(h, k), the vertical line x=hx = h is the axis of symmetry of its parabolic graph.
2
Determine the horizontal distance from the axis of symmetry to the given point x=5x = 5 and the target point x=1x = -1.
The distance for x=5x = 5 is 52=35 - 2 = 3 units. The distance for x=1x = -1 is 2(1)=32 - (-1) = 3 units.
Checking if the two xx-coordinates are symmetric with respect to the line x=2x = 2 allows us to use the symmetry property of parabolas.
3
Apply the symmetry property to find the function value.
Since both x=5x = 5 and x=1x = -1 are equidistant from the axis of symmetry, their function values are equal: f(1)=f(5)=13f(-1) = f(5) = 13.
Points on a parabola that are equidistant from the axis of symmetry have the same yy-coordinate.

Anahtar Kavram

Symmetry of quadratic functions about their vertex axis of symmetry
Soru 53Soru

For the quadratic function ff, the table shows some values of xx and their corresponding values of f(x)f(x).

xxf(x)f(x)
111515
3333
551515

What is the value of f(0)f(0)?

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Cevap: 30

Cevap

30
Since the function values at x=1x = 1 and x=5x = 5 are both 1515, the axis of symmetry of the quadratic function must be located at the midpoint of these values, which is x=3x = 3. The vertex of the function must therefore have an xx-coordinate of 33. From the table, f(3)=3f(3) = 3, which means the vertex is (3,3)(3, 3). The vertex form of the quadratic function is f(x)=a(x3)2+3f(x) = a(x - 3)^2 + 3. To find the value of aa, substitute the point (1,15)(1, 15) into the equation: 15=a(13)2+315 = a(1 - 3)^2 + 3, which simplifies to 15=4a+315 = 4a + 3, giving a=3a = 3. The equation of the function is f(x)=3(x3)2+3f(x) = 3(x - 3)^2 + 3. Substituting x=0x = 0 yields f(0)=3(03)2+3=3(9)+3=30f(0) = 3(0 - 3)^2 + 3 = 3(9) + 3 = 30.

Adım Adım Çözüm

1
Find the axis of symmetry.
The axis of symmetry is x=3x = 3.
Since f(1)=15f(1) = 15 and f(5)=15f(5) = 15, the function values are equal at these points. Because a quadratic function is symmetric, the axis of symmetry is the midpoint of these two xx-values: 1+52=3\frac{1 + 5}{2} = 3.
2
Identify the vertex.
The vertex is (3,3)(3, 3).
The vertex of a quadratic function lies on its axis of symmetry. Since the axis of symmetry is x=3x = 3, the xx-coordinate of the vertex is 33. From the table, f(3)=3f(3) = 3, so the vertex is (3,3)(3, 3).
3
Write the function in vertex form and solve for the leading coefficient.
f(x)=3(x3)2+3f(x) = 3(x - 3)^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. Substituting (3,3)(3, 3), we get f(x)=a(x3)2+3f(x) = a(x - 3)^2 + 3. Using the point (1,15)(1, 15) from the table, we substitute x=1x = 1 and f(x)=15f(x) = 15: 15=a(13)2+315=4a+312=4aa=315 = a(1 - 3)^2 + 3 \Rightarrow 15 = 4a + 3 \Rightarrow 12 = 4a \Rightarrow a = 3.
4
Evaluate the function at x=0x = 0.
f(0)=30f(0) = 30
Substitute x=0x = 0 into the function: f(0)=3(03)2+3=3(9)+3=27+3=30f(0) = 3(0 - 3)^2 + 3 = 3(9) + 3 = 27 + 3 = 30.

Anahtar Kavram

Quadratic Functions and Graphs
Soru 54Soru

In the xyxy-plane, the graph of the quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are constants, has a vertex at (2,3)(2, -3) and passes through the point (4,5)(4, 5). If the graph of a second quadratic function, gg, is obtained by translating the graph of ff horizontally by 33 units to the right and vertically by 55 units up, what is the value of g(5)g(5)?

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Cevap: 2

Cevap

The value of g(5)g(5) is 22.
The vertex of the graph of ff is at (2,3)(2, -3), which means f(2)=3f(2) = -3. The graph of gg is obtained by translating the graph of ff by 33 units to the right and 55 units up, so its equation is g(x)=f(x3)+5g(x) = f(x - 3) + 5. To find g(5)g(5), we substitute x=5x = 5 into this relation, which gives g(5)=f(53)+5=f(2)+5g(5) = f(5 - 3) + 5 = f(2) + 5. Since f(2)=3f(2) = -3, we have g(5)=3+5=2g(5) = -3 + 5 = 2. Alternatively, translating the vertex of ff at (2,3)(2, -3) by 33 units to the right and 55 units up gives the vertex of gg at (2+3,3+5)=(5,2)(2 + 3, -3 + 5) = (5, 2). Since the vertex of the parabola gg occurs at x=5x = 5, the value of g(5)g(5) is the yy-coordinate of the vertex, which is 22.

Adım Adım Çözüm

1
Identify the vertex of the function ff and write its vertex form equation.
The vertex of ff is (2,3)(2, -3), so the vertex form of the function is f(x)=a(x2)23f(x) = a(x - 2)^2 - 3.
This allows us to find the specific equation of f(x)f(x) if needed, and also tells us that the value of f(2)f(2) is 3-3.
2
Set up the equation for the translated function g(x)g(x).
A horizontal translation of 33 units to the right and a vertical translation of 55 units up is represented by g(x)=f(x3)+5g(x) = f(x - 3) + 5.
This defines the function gg in terms of the function ff using standard translation rules.
3
Evaluate g(5)g(5) using the relation from step 2.
g(5)=f(53)+5=f(2)+5g(5) = f(5 - 3) + 5 = f(2) + 5.
This simplifies the calculation by using the value of ff at the vertex.
4
Substitute the value of f(2)f(2) to find g(5)g(5).
g(5)=3+5=2g(5) = -3 + 5 = 2.
Since the vertex of ff is at (2,3)(2, -3), we know f(2)=3f(2) = -3, which yields the final result.

Anahtar Kavram

Quadratic functions can be analyzed and transformed using their vertex form and function translation rules.
Tahmini Süre:1m 30s
Soru 55Soru

The graph of the quadratic function f(x)=x26x+cf(x) = x^2 - 6x + c, where cc is a constant, has its vertex at (h,k)(h, k) in the xyxy-plane. If the graph of ff is translated 33 units to the right and 22 units down, the vertex of the translated graph lies on the line y=2xy = 2x. What is the value of cc?

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Cevap: 23

Cevap

23
To find the constant cc, we first determine the vertex of the function f(x)=x26x+cf(x) = x^2 - 6x + c. The x-coordinate of the vertex of a parabola in the form y=ax2+bx+cy = ax^2 + bx + c is given by x=b2ax = -\frac{b}{2a}. For this function, x=62(1)=3x = -\frac{-6}{2(1)} = 3. Substituting this back into the function gives the y-coordinate of the vertex: f(3)=(3)26(3)+c=c9f(3) = (3)^2 - 6(3) + c = c - 9. Thus, the original vertex is at (3,c9)(3, c - 9). Translating the graph 33 units to the right increases the x-coordinate of the vertex by 33, making it 3+3=63 + 3 = 6. Translating the graph 22 units down decreases the y-coordinate of the vertex by 22, making it (c9)2=c11(c - 9) - 2 = c - 11. The problem states that this new vertex (6,c11)(6, c - 11) lies on the line y=2xy = 2x. Substituting these coordinates into the linear equation gives c11=2(6)c - 11 = 2(6), which simplifies to c11=12c - 11 = 12. Adding 1111 to both sides gives the value of cc as 2323.

Adım Adım Çözüm

1
Find the vertex (h,k)(h, k) of the original quadratic function f(x)=x26x+cf(x) = x^2 - 6x + c.
The vertex is at (3,c9)(3, c - 9).
The x-coordinate of the vertex of a quadratic function y=ax2+bx+cy = ax^2 + bx + c is given by h=b2ah = -\frac{b}{2a}. For f(x)=x26x+cf(x) = x^2 - 6x + c, we have h=62(1)=3h = -\frac{-6}{2(1)} = 3. Substituting x=3x = 3 into the function gives the y-coordinate: k=f(3)=326(3)+c=c9k = f(3) = 3^2 - 6(3) + c = c - 9.
2
Determine the coordinates of the vertex after translating the graph 33 units to the right and 22 units down.
The new vertex is at (6,c11)(6, c - 11).
A translation of 33 units to the right increases the x-coordinate of the vertex by 33, so the new x-coordinate is 3+3=63 + 3 = 6. A translation of 22 units down decreases the y-coordinate of the vertex by 22, so the new y-coordinate is (c9)2=c11(c - 9) - 2 = c - 11.
3
Set up an equation using the line y=2xy = 2x and solve for cc.
c=23c = 23
Since the translated vertex (6,c11)(6, c - 11) lies on the line y=2xy = 2x, substituting x=6x = 6 and y=c11y = c - 11 into the line's equation must satisfy it: c11=2(6)c11=12c=23c - 11 = 2(6) \Rightarrow c - 11 = 12 \Rightarrow c = 23.

Anahtar Kavram

Determining the vertex of a quadratic function and applying translations to its graph.
Tahmini Süre:1m 30s
Soru 56Soru

The graph of the quadratic function ff in the xyxy-plane has its vertex at (4,12)(4, 12). The function is defined by f(x)=(xc)(xd)f(x) = -(x - c)(x - d), where cc and dd are constants. What is the value of the product cdcd?

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Cevap: 4

Cevap

The value of the product cdcd is 44.
The xx-coordinate of the vertex of a quadratic function of the form f(x)=(xc)(xd)f(x) = -(x-c)(x-d) is the average of the xx-intercepts cc and dd. Since the vertex is (4,12)(4, 12), we have c+d2=4\frac{c+d}{2} = 4, which means c+d=8c+d = 8. Substituting the vertex (4,12)(4, 12) into the function gives 12=(4c)(4d)12 = -(4-c)(4-d), which expands to 12=(164(c+d)+cd)12 = -(16 - 4(c+d) + cd). Substituting c+d=8c+d = 8 gives 12=(1632+cd)=(16+cd)=16cd12 = -(16 - 32 + cd) = -(-16 + cd) = 16 - cd. Solving for cdcd yields cd=4cd = 4.

Adım Adım Çözüm

1
Find the sum of the constants cc and dd using the xx-coordinate of the vertex.
c+d=8c + d = 8
The graph of a quadratic function in the form f(x)=(xc)(xd)f(x) = -(x-c)(x-d) has a vertical line of symmetry at the xx-coordinate of its vertex, which is the midpoint of its xx-intercepts cc and dd. Therefore, c+d2=4\frac{c+d}{2} = 4, which simplifies to c+d=8c + d = 8.
2
Substitute the vertex coordinates (4,12)(4, 12) into the function definition.
(4c)(4d)=12(4-c)(4-d) = -12
Since (4,12)(4, 12) is the vertex, the point lies on the graph of ff, meaning f(4)=12f(4) = 12. Substituting x=4x = 4 into the function gives 12=(4c)(4d)12 = -(4-c)(4-d), which simplifies to (4c)(4d)=12(4-c)(4-d) = -12.
3
Expand the expression (4c)(4d)(4-c)(4-d) and substitute c+d=8c+d = 8 to solve for cdcd.
cd=4cd = 4
Expanding (4c)(4d)=12(4-c)(4-d) = -12 gives 164(c+d)+cd=1216 - 4(c+d) + cd = -12. Substituting c+d=8c+d = 8 yields 1632+cd=1216 - 32 + cd = -12, which simplifies to 16+cd=12-16 + cd = -12. Adding 1616 to both sides gives cd=4cd = 4.

Anahtar Kavram

Quadratic Functions and Graphs
Soru 57Soru

The graph of the quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are constants, is a parabola in the xyxy-plane that passes through the points (3,22)(-3, 22) and (9,22)(9, 22). If the minimum value of f(x)f(x) is 44, what is the value of f(1)f(1)?

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Cevap: 6

Cevap

The correct answer is 6.
The correct answer is 6. The axis of symmetry of the parabola is halfway between the points with equivalent y-values: x = 3. Using the minimum value of 4, the vertex is identified as (3, 4). Writing the equation in vertex form as f(x) = a(x - 3)^2 + 4 and substituting (9, 22) yields a = 0.5. Evaluating the function f(x) = 0.5(x - 3)^2 + 4 at x = 1 yields 6.

Adım Adım Çözüm

1
Determine the axis of symmetry of the parabola.
The axis of symmetry is x=3x = 3.
Because the parabola passes through the points (3,22)(-3, 22) and (9,22)(9, 22), which have the same yy-coordinate, the axis of symmetry must lie halfway between their xx-coordinates: x=3+92=3x = \frac{-3 + 9}{2} = 3.
2
Write the quadratic function in vertex form.
f(x)=a(x3)2+4f(x) = a(x - 3)^2 + 4
Since the function has a minimum value of 44, the vertex of the upward-opening parabola is at (3,4)(3, 4).
3
Solve for the leading coefficient aa.
a=0.5a = 0.5
Substitute the point (9,22)(9, 22) into the vertex form equation: 22=a(93)2+422 = a(9 - 3)^2 + 4, which simplifies to 18=36a18 = 36a, so a=0.5a = 0.5.
4
Evaluate f(1)f(1).
f(1)=6f(1) = 6
Substitute x=1x = 1 into the completed function f(x)=0.5(x3)2+4f(x) = 0.5(x - 3)^2 + 4 to get f(1)=0.5(13)2+4=0.5(4)+4=6f(1) = 0.5(1 - 3)^2 + 4 = 0.5(4) + 4 = 6.

Anahtar Kavram

Using symmetry, vertex form, and given points to determine a quadratic function's equation and evaluate it.
Soru 58Soru

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?

Cevabı ve açıklamayı göster

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 59Soru

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?

Cevabı ve açıklamayı göster

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 60Soru

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?

Cevabı ve açıklamayı göster

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
ÖncekiSayfa 3 / 4Sonraki
Quadratic Functions and Graphs Alıştırma Soruları — SAT — Sayfa 3 | Examkin