Advanced Math Alıştırma Soruları — SAT — Sayfa 21

The table below shows selected values for a polynomial function $f$ of degree 3.

$x$	$f(x)$
$-2$	$0$
$1$	$0$
$3$	$0$
$4$	$18$

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

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Cevap: 6

Cevap

The value of

f(0)

is 6.

The table shows the roots of the cubic function

f(x)

are

x = -2

,

x = 1

, and

x = 3

. This allows the function to be written in factored form as

f(x) = a(x + 2)(x - 1)(x - 3)

. Using the point

(4, 18)

from the table, we solve for

a

:

18 = a(4 + 2)(4 - 1)(4 - 3) \Rightarrow 18 = 18a \Rightarrow a = 1

. The function is

f(x) = (x + 2)(x - 1)(x - 3)

. Substituting

x = 0

gives

f(0) = (2)(-1)(-3) = 6

.

Adım Adım Çözüm

1

Identify the roots of the polynomial from the table.

The roots of the function are

x = -2

,

x = 1

, and

x = 3

.

The table shows that

f(x) = 0

at these values, indicating they are the

x

-intercepts or roots.

2

Write the general factored form of a degree 3 polynomial with these roots.

f(x) = a(x + 2)(x - 1)(x - 3)

A cubic polynomial with roots

r_1

,

r_2

, and

r_3

can be expressed as

f(x) = a(x - r_1)(x - r_2)(x - r_3)

for a constant coefficient

a

.

3

Find the value of the constant coefficient

a

using a known point from the table.

a = 1

Substituting the table values

x = 4

and

f(4) = 18

yields

18 = a(4 + 2)(4 - 1)(4 - 3)

, which simplifies to

18 = 18a

, so

a = 1

.

4

Calculate the value of

f(0)

using the fully defined function.

f(0) = 6

Substituting

x = 0

into the equation

f(x) = (x + 2)(x - 1)(x - 3)

gives

f(0) = (0 + 2)(0 - 1)(0 - 3) = 2 \times (-1) \times (-3) = 6

.

Anahtar Kavram

Using polynomial roots and extra points to define a polynomial function and evaluate it.

Soru 402Soru →

For the function $f(x) = 4^x - 3$ , a new function $g$ is created by reflecting the graph of $y = f(x)$ across the $x$ -axis, followed by a vertical translation of $c$ units upward. The graph of $y = g(x)$ contains the point $(2, -5)$ . What is the value of the constant $c$ ?

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Cevap: 8

Cevap

8

Reflecting the graph of

f(x) = 4^x - 3

across the

x

-axis results in the function

-f(x) = -(4^x - 3) = -4^x + 3

. Translating this graph vertically upward by

c

units defines

g(x) = -4^x + 3 + c

. Substituting the given point

(2, -5)

yields

-5 = -4^2 + 3 + c

, which simplifies to

-5 = -13 + c

. Adding

13

to both sides gives

c = 8

.

Adım Adım Çözüm

1

Reflect the function

f(x)

across the

x

-axis

-f(x) = -(4^x - 3) = -4^x + 3

Reflecting a graph across the

x

-axis is represented by negating the output of the function, which transforms

y = f(x)

to

y = -f(x)

.

2

Apply the vertical translation upward by

c

units to define

g(x)

g(x) = -f(x) + c = -4^x + 3 + c

Translating a function vertically upward by

c

units adds

c

to the function's expression.

3

Substitute the point

(2, -5)

into

g(x)

and solve for

c

c = 8

Since the graph of

g

passes through

(2, -5)

, substitute

x = 2

and

g(x) = -5

into the equation:

-5 = -4^2 + 3 + c

. This simplifies to

-5 = -16 + 3 + c

, then

-5 = -13 + c

, which gives

c = 8

.

Anahtar Kavram

Function transformations, including vertical reflections and translations

Soru 403Soru →

A polynomial function $p$ is defined by $p(x) = ax(x-3)^2$ , where $a$ is a constant. In the $xy$ -plane, the graph of $y = p(x)$ passes through the point $(1, -8)$ . What is the value of $p(2)$ ?

-25

-8

-4

4

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Cevap: -4

Cevap

-4

The correct answer is -4. By substituting the point

(1, -8)

into the given function

p(x) = ax(x-3)^2

, we obtain

-8 = a(1)(1-3)^2

, which simplifies to

-8 = 4a

. Dividing both sides by

4

gives

a = -2

. Substituting

a = -2

back into the function yields

p(x) = -2x(x-3)^2

. Evaluating this function at

x = 2

gives

p(2) = -2(2)(2-3)^2 = -4(1) = -4

.

Adım Adım Çözüm

1

Substitute the coordinates of the given point

(1, -8)

into the polynomial equation to set up an equation for the constant

a

.

p(1) = a(1)(1 - 3)^2 = -8

Since the graph passes through

(1, -8)

, substituting

x = 1

must yield

y = -8

.

2

Simplify the expression and solve for the constant

a

.

a(-2)^2 = -8 \Rightarrow 4a = -8 \Rightarrow a = -2

Evaluating

(1-3)^2

gives

4

, leading to a simple linear equation in terms of

a

.

3

Substitute

a = -2

back into the original polynomial definition to obtain the complete function formula.

p(x) = -2x(x - 3)^2

Knowing the value of the leading constant allows us to write the explicit formula for

p(x)

.

4

Evaluate

p(2)

by substituting

x = 2

into the completed function formula.

p(2) = -2(2)(2 - 3)^2 = -4(-1)^2 = -4

This calculation yields the value of the polynomial at

x = 2

as requested.

Anahtar Kavram

Polynomial Factors and Graphs

Soru 404Soru →

In the $xy$ -plane, the graph of the quadratic function $f$ is defined by $f(x) = -x^2 + 6x - 5$ . If the graph of $f$ is translated $2$ units to the right and $4$ units up to form the graph of a new function $g$ , what is the maximum value of $g$ ?

0

5

8

10

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

The maximum value of the quadratic function is the y-coordinate of its vertex. For

f(x) = -x^2 + 6x - 5

, the vertex occurs at

x = -\frac{6}{2(-1)} = 3

. Evaluating

f(3)

yields

-9 + 18 - 5 = 4

, which is the maximum value of

f

. Translating the graph

2

units to the right shifts the vertex horizontally to

x = 5

but does not change the maximum height. Translating the graph

4

units up increases the y-coordinate of the vertex by

4

, resulting in a new maximum value of

4 + 4 = 8

.

Adım Adım Çözüm

1

Find the x-coordinate of the vertex of the function

f(x) = -x^2 + 6x - 5

.

x = 3

The x-coordinate of the vertex of a quadratic function in the standard form

y = ax^2 + bx + c

is given by

x = -\frac{b}{2a}

. For

f(x)

, we have

a = -1

and

b = 6

, so

x = -\frac{6}{2(-1)} = 3

.

2

Calculate the maximum value of

f(x)

by finding the y-coordinate of its vertex.

f(3) = 4

Substitute

x = 3

into the function:

f(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4

. Since the coefficient of

x^2

is negative, the parabola opens downward, meaning the vertex represents the maximum value.

3

Apply the vertical translation to find the maximum value of the new function

g(x)

.

Maximum value of

g

is

8

Translating the graph

2

units to the right shifts the vertex horizontally but does not affect the vertical maximum value. Translating the graph

4

units up increases the maximum value by

4

. Thus, the new maximum value is

4 + 4 = 8

.

Anahtar Kavram

Determining the maximum value of a quadratic function and applying translations to its graph.

Soru 405Soru →

The function $f$ is defined by $f(x) = -(\sqrt{x+2} - 1)$ . The graph of the function $g$ in the $xy$ -plane is obtained by shifting the graph of $f$ to the right by $3$ units and then down by $2$ units. If the point $(a, -5)$ lies on the graph of $g$ , what is the value of $a$ ?

5

11

17

37

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Cevap: 17

Cevap

17

The correct answer is

17

. Shifting the graph of

f(x) = -(\sqrt{x+2} - 1)

to the right by

3

units is represented by

f(x-3) = -(\sqrt{x-1} - 1) = -\sqrt{x-1} + 1

. Shifting this result down by

2

units gives the function

g(x) = f(x-3) - 2 = -\sqrt{x-1} - 1

. Substituting the point

(a, -5)

into the equation for

g(x)

yields

-\sqrt{a-1} - 1 = -5

. Adding

1

to both sides gives

-\sqrt{a-1} = -4

, or

\sqrt{a-1} = 4

. Squaring both sides results in

a - 1 = 16

. Finally, adding

1

to both sides gives

a = 17

.

Adım Adım Çözüm

1

Apply the horizontal translation to the function

f(x)

.

f(x-3) = -(\sqrt{x-3+2} - 1) = -(\sqrt{x-1} - 1) = -\sqrt{x-1} + 1

Shifting a function

f(x)

to the right by

3

units replaces

x

with

x-3

.

2

Apply the vertical translation to find the expression for

g(x)

.

g(x) = f(x-3) - 2 = -\sqrt{x-1} + 1 - 2 = -\sqrt{x-1} - 1

Shifting a graph down by

2

units subtracts

2

from the entire function expression.

3

Substitute the point

(a, -5)

into the equation for

g(x)

and solve for

a

.

-\sqrt{a-1} - 1 = -5 \implies -\sqrt{a-1} = -4 \implies \sqrt{a-1} = 4 \implies a-1 = 16 \implies a = 17

Since the point

(a, -5)

lies on the graph of

g

, we set

g(a) = -5

and solve for

a

by isolating the radical and squaring both sides.

Anahtar Kavram

Function transformations involve substituting

x-h

for

x

for horizontal translations and adding

k

to the function for vertical translations.

Tahmini Süre:1m 30s

Soru 406Soru →

A projectile is launched from a platform. The height $h(t)$ , in meters, of the projectile $t$ seconds after launch is modeled by the equation $h(t) = -5t^2 + v_0 t + h_0$ , where $v_0$ and $h_0$ are constants. If the projectile reaches its maximum height of $80$ meters at $3$ seconds after launch, what is the value of $h_0$ ?

5

35

80

125

Cevabı ve açıklamayı göster

Cevap: 35

Cevap

The value of

h_0

is

35

, which represents the initial height of the projectile.

The maximum height of

80

meters at

3

seconds indicates that the vertex of the quadratic function is

(3, 80)

. Since the lead coefficient of the

t^2

term is

-5

, we can express the function in vertex form as

h(t) = -5(t - 3)^2 + 80

. To find the value of

h_0

, which represents the height at

t = 0

, we substitute

0

for

t

, yielding

h(0) = -5(0 - 3)^2 + 80 = -5(9) + 80 = -45 + 80 = 35

. Thus, the initial height is

35

meters.

Adım Adım Çözüm

1

Identify the coordinates of the vertex of the parabola from the context.

The maximum height is

80

meters at

3

seconds, so the vertex of the function is

(3, 80)

.

The vertex of a downward-opening parabola represents its maximum value.

2

Write the quadratic function in vertex form using the vertex

(3, 80)

and the given leading coefficient

a = -5

.

h(t) = -5(t - 3)^2 + 80

The vertex form of a quadratic equation is

h(t) = a(t - d)^2 + k

, where

(d, k)

is the vertex.

3

Evaluate the function at

t = 0

to find the initial height

h_0

.

h(0) = -5(0 - 3)^2 + 80 = -5(9) + 80 = -45 + 80 = 35

The constant term

h_0

in standard form

h(t) = -5t^2 + v_0 t + h_0

represents the height when

t = 0

.

Anahtar Kavram

Vertex of a quadratic function and converting between vertex form and standard form.

Soru 407Soru →

The graph of the quadratic equation $y = 2x^2 - 12x + c$ , where $c$ is a constant, is a parabola in the $xy$ -plane. If the $y$ -coordinate of the vertex of this parabola is $-5$ , what is the value of $c$ ?

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Cevap: 13

Cevap

13

To find the value of the constant

c

, we calculate the coordinates of the vertex of the parabola. The x-coordinate of the vertex for a quadratic function in standard form

y = ax^2 + bx + c

is given by

x = -\frac{b}{2a}

. Substituting

a = 2

and

b = -12

gives

x = -\frac{-12}{2(2)} = 3

. Evaluating the quadratic equation at

x = 3

gives the y-coordinate of the vertex:

y = 2(3)^2 - 12(3) + c = 18 - 36 + c = c - 18

. Since we are given that the y-coordinate of the vertex is

-5

, we set

c - 18 = -5

and solve to find

c = 13

.

Adım Adım Çözüm

1

Calculate the x-coordinate of the vertex using the vertex formula.

x = 3

The axis of symmetry and the x-coordinate of the vertex are located at

x = -\frac{b}{2a}

.

2

Substitute the x-coordinate of the vertex into the equation to express the y-coordinate in terms of

c

.

y = c - 18

Evaluating the quadratic function at the vertex's x-coordinate gives the minimum or maximum value of the function.

3

Equate the expression for the y-coordinate to the given vertex y-coordinate of

-5

and solve for

c

.

c = 13

Setting the calculated y-coordinate expression equal to the given value allows us to isolate and solve for the unknown constant.

Anahtar Kavram

Determining the vertex of a quadratic function from its standard form and solving for a constant coefficient.

Alternatif Yöntem

Alternatively, we can complete the square to write the quadratic equation in vertex form,

y = a(x-h)^2 + k

. Factoring the leading coefficient from the variable terms gives

y = 2(x^2 - 6x) + c

. To complete the square inside the parentheses, add and subtract

9

:

y = 2(x^2 - 6x + 9 - 9) + c = 2((x-3)^2 - 9) + c = 2(x-3)^2 - 18 + c

. In this vertex form, the y-coordinate of the vertex is

k = c - 18

. Since the vertex y-coordinate is

-5

, we set

c - 18 = -5

to get

c = 13

.

Tahmini Süre:1m 30s

Soru 408Soru →

In the quadratic equation $x^2 - kx + 5 = 0$ , $k$ is a constant. If the solutions to the equation are $x = 3 \pm \sqrt{a}$ , where $a$ is a positive constant, what is the value of $a$ ?

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Cevap: 4

Cevap

4

The correct answer is 4. By relating the roots of a quadratic equation to its coefficients, the sum of the roots is

(3 + \sqrt{a}) + (3 - \sqrt{a}) = 6

, which equals the linear coefficient term

k

in

x^2 - kx + 5 = 0

. Using

k = 6

, we can rewrite the equation as

x^2 - 6x + 5 = 0

. Completing the square yields

(x - 3)^2 - 9 + 5 = 0

, which simplifies to

(x - 3)^2 = 4

, or

x = 3 \pm \sqrt{4}

. Comparing this to the given expression

x = 3 \pm \sqrt{a}

gives

a = 4

. Alternatively, using the product of roots,

(3 + \sqrt{a})(3 - \sqrt{a}) = 9 - a

. Since the product of roots is the constant term 5, we have

9 - a = 5

, which directly yields

a = 4

.

Adım Adım Çözüm

1

Find the sum of the given solutions to determine the value of the constant

k

.

k = 6

The sum of the solutions

(3 + \sqrt{a}) + (3 - \sqrt{a})

is

6

. For a quadratic equation in the form

x^2 - kx + c = 0

, the sum of the solutions is equal to

k

.

2

Substitute

k = 6

back into the quadratic equation.

x^2 - 6x + 5 = 0

This gives the complete quadratic equation with numerical coefficients.

3

Solve the quadratic equation by completing the square to express the solutions in the form

3 \pm \sqrt{a}

.

x = 3 \pm \sqrt{4}

Completing the square gives

(x - 3)^2 - 9 + 5 = 0

, which simplifies to

(x - 3)^2 = 4

, so

x - 3 = \pm \sqrt{4}

.

4

Compare the obtained solutions to the given form

x = 3 \pm \sqrt{a}

to identify the value of

a

.

a = 4

Comparing

x = 3 \pm \sqrt{4}

with

x = 3 \pm \sqrt{a}

yields

a = 4

.

Anahtar Kavram

Relationship between roots and coefficients of a quadratic equation, or solving quadratic equations by completing the square.

Soru 409Soru →

The quadratic function $f$ is defined by $f(x) = (x - 3)(x - 7)$ . The function $g$ is defined by $g(x) = f(x - 2) + k$ , where $k$ is a constant. In the $xy$ -plane, the vertex of the graph of $y = g(x)$ is $(7, 5)$ . What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

9

To find the vertex of the function

f(x) = (x - 3)(x - 7)

, we calculate the midpoint of the

x

-intercepts,

x = 5

, and evaluate the function at this value to get

f(5) = -4

, giving the vertex

(5, -4)

. The graph of

g(x) = f(x - 2) + k

shifts the graph of

f

right by

2

units and up by

k

units, resulting in a vertex of

(7, -4 + k)

. Since the vertex of

g(x)

is

(7, 5)

, we have

-4 + k = 5

, which solves to

k = 9

.

Adım Adım Çözüm

1

Find the vertex of the quadratic function

f(x) = (x - 3)(x - 7)

.

The vertex of

f(x)

is

(5, -4)

.

The

x

-coordinate of the vertex of a quadratic function in factored form

f(x) = a(x - r_1)(x - r_2)

lies midway between the roots

r_1

and

r_2

. Here,

x = \frac{3 + 7}{2} = 5

. Evaluating the function at

x = 5

gives the

y

-coordinate:

f(5) = (5 - 3)(5 - 7) = -4

.

2

Apply the transformations to find the coordinates of the vertex of

g(x) = f(x - 2) + k

.

The vertex of

g(x)

is

(7, -4 + k)

.

The transformation

g(x) = f(x - h) + k

shifts the graph of

f(x)

horizontally by

h

units to the right and vertically by

k

units up. For

g(x) = f(x - 2) + k

, the vertex of

f

,

(5, -4)

, is shifted

2

units to the right and

k

units up, resulting in the new vertex

(5 + 2, -4 + k) = (7, -4 + k)

.

3

Equate the

y

-coordinate of the vertex of

g(x)

to the given value and solve for

k

.

k = 9

We are given that the vertex of the graph of

y = g(x)

is

(7, 5)

. Setting the

y

-coordinate of the transformed vertex equal to

5

gives

-4 + k = 5

. Adding

4

to both sides yields

k = 9

.

Anahtar Kavram

Function Transformations of Quadratic Graphs

Soru 410Soru →

The graph of a third-degree polynomial function $g$ has $x$ -intercepts at $(-2, 0)$ and $(3, 0)$ . At $x = -2$ , the graph is tangent to the $x$ -axis, and the graph passes through the point $(1, -12)$ . What is the $y$ -intercept of the graph of $g$ ?

(0, -12)

(0, -8)

(0, -18)

(0, -36)

Cevabı ve açıklamayı göster

Cevap: $(0, -8)$

Cevap

The

y

-intercept of the graph is

(0, -8)

.

The polynomial has a double root at

x = -2

because it is tangent to the

x

-axis at

(-2, 0)

, and a single root at

x = 3

because it crosses the

x

-axis at

(3, 0)

. This allows us to write the cubic polynomial as

g(x) = a(x+2)^2(x-3)

. Substituting the point

(1, -12)

gives

-12 = a(1+2)^2(1-3) \implies -12 = -18a \implies a = \frac{2}{3}

. To find the

y

-intercept, we evaluate the function at

x = 0

:

g(0) = \frac{2}{3}(0+2)^2(0-3) = \frac{2}{3}(4)(-3) = -8

, which corresponds to the point

(0, -8)

.

Adım Adım Çözüm

1

Set up the general form of the cubic function using the given intercepts.

g(x) = a(x+2)^2(x-3)

An

x

-intercept at

(-2, 0)

where the graph is tangent to the

x

-axis represents a root of multiplicity 2, which corresponds to the factor

(x+2)^2

. An

x

-intercept at

(3, 0)

represents a root of multiplicity 1, corresponding to the factor

(x-3)

.

2

Use the point

(1, -12)

to solve for the constant coefficient

a

.

a = \frac{2}{3}

Substitute

x = 1

and

g(1) = -12

into the equation:

-12 = a(1+2)^2(1-3) \implies -12 = a(9)(-2) \implies -12 = -18a \implies a = \frac{2}{3}

.

3

Calculate the

y

-intercept of the function by evaluating

g(0)

.

g(0) = -8

, which corresponds to the point

(0, -8)

The

y

-intercept of a graph occurs where

x = 0

. Substituting

x = 0

into the equation

g(x) = \frac{2}{3}(x+2)^2(x-3)

yields

g(0) = \frac{2}{3}(0+2)^2(0-3) = \frac{2}{3}(4)(-3) = -8

.

Anahtar Kavram

Identifying polynomial factors and equations from graphical features like intercepts and tangency.

Soru 411Soru →

The graph of the quadratic function $h$ has a vertex at $(2, -5)$ in the $xy$ -plane. If the function $g$ is defined by $g(x) = h(x + 3) + 4$ , what is the vertex of the graph of $g$ ?

(-1, -1)

(5, -1)

(-5, -1)

(1, -1)

Cevabı ve açıklamayı göster

Cevap: $(-1, -1)$

Cevap

The vertex of the graph of

g

is

(-1, -1)

.

The vertex of the graph of

h

is given as

(2, -5)

. The function

g(x) = h(x + 3) + 4

represents a transformation of the graph of

h

. Specifically,

h(x + 3)

shifts the graph horizontally to the left by

3

units, and adding

4

shifts the graph vertically up by

4

units. Applying these shifts to the original vertex coordinates: the new

x

-coordinate is

2 - 3 = -1

, and the new

y

-coordinate is

-5 + 4 = -1

. Therefore, the vertex of the graph of

g

is

(-1, -1)

.

Adım Adım Çözüm

1

Identify the vertex of the original function

h

.

The vertex of

h

is

(2, -5)

.

The problem states that the graph of the quadratic function

h

has a vertex at

(2, -5)

.

2

Determine the horizontal translation from the expression

h(x + 3)

.

A horizontal shift of

3

units to the left.

Replacing

x

with

x + c

in a function

f(x)

shifts the graph left by

c

units. Here,

c = 3

, so we shift the vertex

x

-coordinate left by

3

:

2 - 3 = -1

.

3

Determine the vertical translation from the addition of

4

.

A vertical shift of

4

units up.

Adding a constant

d

to a function shifts its graph vertically by

d

units. Here,

d = 4

, so we shift the vertex

y

-coordinate up by

4

:

-5 + 4 = -1

.

4

Combine the shifted coordinates to find the new vertex.

The new vertex is

(-1, -1)

.

Combining the translated

x

-coordinate of

-1

and

y

-coordinate of

-1

gives the vertex of the graph of

g

.

Anahtar Kavram

Quadratic Function Transformations

Soru 412Soru →

The function $f$ is defined by $f(x) = (x - 2)^2 - 5$ . The graph of the function $g$ in the $xy$ -plane is obtained by translating the graph of $f$ by $3$ units to the left and $4$ units up. If the vertex of the graph of $g$ is $(h, k)$ , what is the value of $h + k$ ?

-6

-2

4

-10

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

-2

The correct answer is

-2

. The vertex form of a quadratic function is given by

y = a(x - h)^2 + k

, where

(h, k)

is the vertex. For the function

f(x) = (x - 2)^2 - 5

, the vertex is at

(2, -5)

. Translating the graph of

f

by

3

units to the left shifts the

x

-coordinate of the vertex to

2 - 3 = -1

. Translating the graph of

f

by

4

units up shifts the

y

-coordinate of the vertex to

-5 + 4 = -1

. Therefore, the vertex of the graph of

g

is

(h, k) = (-1, -1)

, and the value of

h + k

is

-1 + (-1) = -2

.

Adım Adım Çözüm

1

Identify the vertex of the original function

f(x) = (x - 2)^2 - 5

.

The vertex is

(2, -5)

.

For a quadratic function in vertex form

y = a(x - h)^2 + k

, the coordinates of the vertex are

(h, k)

.

2

Apply the horizontal translation of

3

units to the left to the vertex's

x

-coordinate.

The new

x

-coordinate is

2 - 3 = -1

.

Shifting a graph horizontally to the left by

c

units subtracts

c

from the

x

-coordinates of its points.

3

Apply the vertical translation of

4

units up to the vertex's

y

-coordinate.

The new

y

-coordinate is

-5 + 4 = -1

.

Shifting a graph vertically upward by

c

units adds

c

to the

y

-coordinates of its points.

4

Calculate the sum of the coordinates of the new vertex

(h, k) = (-1, -1)

.

h + k = -1 + (-1) = -2

.

The question asks for the value of the sum

h + k

, where

(h, k)

is the vertex of the graph of

g

.

Anahtar Kavram

Quadratic function vertex transformations

Soru 413Soru →

The graph of the polynomial function $h$ in the $xy$ -plane has $x$ -intercepts only at $(-4, 0)$ , $(1, 0)$ , and $(3, 0)$ . If the $y$ -intercept of the graph of $h$ is $(0, -12)$ and the degree of $h$ is $3$ , what is the value of $h(2)$ ?

6

-6

-30

30

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6

Since the degree of the polynomial function

h

is 3 and its only

x

-intercepts are

(-4, 0)

,

(1, 0)

, and

(3, 0)

, the function can be expressed in factored form as

h(x) = a(x + 4)(x - 1)(x - 3)

, where

a

is a constant. We can find the value of

a

by using the

y

-intercept of

(0, -12)

. Substituting

x = 0

into the equation gives

-12 = a(0 + 4)(0 - 1)(0 - 3)

, which simplifies to

-12 = 12a

, so

a = -1

. The complete function is therefore

h(x) = -(x + 4)(x - 1)(x - 3)

. To find

h(2)

, substitute

2

for

x

:

h(2) = -(2 + 4)(2 - 1)(2 - 3) = -(6)(1)(-1) = 6

. This matches the correct answer.

Adım Adım Çözüm

1

Write the general form of the cubic polynomial function

h(x)

using its

x

-intercepts.

h(x) = a(x + 4)(x - 1)(x - 3)

Since the degree of

h

is 3 and its only

x

-intercepts are

(-4, 0)

,

(1, 0)

, and

(3, 0)

, the corresponding factors must be

(x - (-4)) = (x + 4)

,

(x - 1)

, and

(x - 3)

, multiplied by a constant leading coefficient

a

.

2

Use the

y

-intercept

(0, -12)

to solve for the constant coefficient

a

.

a = -1

Substituting

x = 0

and

h(0) = -12

into the equation gives

-12 = a(0 + 4)(0 - 1)(0 - 3)

, which simplifies to

-12 = 12a

. Solving for

a

gives

a = -1

.

3

Evaluate

h(2)

using the fully defined function

h(x) = -(x + 4)(x - 1)(x - 3)

.

h(2) = 6

Substitute

x = 2

into the polynomial:

h(2) = -(2 + 4)(2 - 1)(2 - 3) = -(6)(1)(-1) = 6

.

Anahtar Kavram

Polynomial Factors and Graphs

Tahmini Süre:1m 30s

Soru 414Soru →

The polynomial function $p$ is defined by $p(x) = x^4 - 3x^3 + ax^2 - 7x + 10$ , where $a$ is a constant. If $x - 2$ is a factor of $p(x)$ , what is the value of $a$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

According to the Factor Theorem, if

x - 2

is a factor of the polynomial function

p(x)

, then evaluating the polynomial at

x = 2

must yield a value of

0

. Substituting

x = 2

into the equation gives

p(2) = 2^4 - 3(2)^3 + a(2)^2 - 7(2) + 10 = 0

. Simplifying the terms results in

16 - 24 + 4a - 14 + 10 = 0

, which further simplifies to

4a - 12 = 0

. Adding

12

to both sides and dividing by

4

yields

a = 3

.

Adım Adım Çözüm

1

Apply the Factor Theorem to set up the equation.

p(2) = 0

By the Factor Theorem,

x - c

is a factor of

p(x)

if and only if

p(c) = 0

.

2

Substitute

x = 2

into the polynomial expression.

2^4 - 3(2)^3 + a(2)^2 - 7(2) + 10 = 0

Evaluating

p(2)

yields a linear equation in terms of the unknown constant

a

.

3

Simplify the arithmetic terms in the equation.

4a - 12 = 0

Simplifying the constant values:

16 - 24 - 14 + 10 = -12

.

4

Solve the linear equation for

a

.

a = 3

Isolating the variable term gives

4a = 12

, and dividing by

4

yields

a = 3

.

Anahtar Kavram

The Factor Theorem states that a polynomial

p(x)

has a factor

x - c

if and only if

p(c) = 0

.

Soru 415Soru →

A quadratic function $f$ is defined by $f(x) = a(x - 4)(x - k)$ , where $a$ and $k$ are constants. In the $xy$ -plane, the graph of $y = f(x)$ has its vertex at $(6, 12)$ . What is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The value of

k

is

8

.

The given quadratic function is in the factored form

f(x) = a(x - 4)(x - k)

, meaning the x-intercepts of its graph are

x = 4

and

x = k

. Due to the symmetry of a parabola, the x-coordinate of the vertex is the midpoint of the x-intercepts. Since the vertex is at

(6, 12)

, its x-coordinate is

6

. Setting the midpoint of the intercepts equal to 6 gives the equation

\frac{4 + k}{2} = 6

. Multiplying both sides by 2 gives

4 + k = 12

, and subtracting 4 gives the correct answer

k = 8

.

Adım Adım Çözüm

1

Identify the x-intercepts of the function

f(x) = a(x - 4)(x - k)

.

The x-intercepts are at

x = 4

and

x = k

.

For any quadratic function written in the factored form

f(x) = a(x - r_1)(x - r_2)

, the values

r_1

and

r_2

correspond to the x-intercepts of the graph.

2

Relate the x-coordinate of the vertex to the x-intercepts using symmetry.

The axis of symmetry equation is

\frac{4 + k}{2} = 6

.

Because a parabola is symmetric, the x-coordinate of the vertex always lies exactly halfway between the two x-intercepts.

3

Solve the equation for

k

.

k = 8

.

Multiplying both sides of the equation

\frac{4 + k}{2} = 6

by 2 yields

4 + k = 12

. Subtracting 4 from both sides gives

k = 8

.

Anahtar Kavram

Symmetry of quadratic graphs and their vertices relative to their x-intercepts.

Alternatif Yöntem

Alternatively, substitute the vertex coordinates

(6, 12)

into the function:

12 = a(6 - 4)(6 - k) = 2a(6 - k)

. Since the vertex is the maximum point, the derivative

f'(x) = a(x - k) + a(x - 4)

must be equal to 0 at

x = 6

. This yields

a(6 - k) + a(6 - 4) = 0

. Since

a \neq 0

, dividing by

a

gives

(6 - k) + 2 = 0

, which simplifies to

8 - k = 0

, or

k = 8

.

Tahmini Süre:1m 30s

Soru 416Soru →

The graph of the quadratic function $f$ in the $xy$ -plane is a parabola. If $f(-3) = -2$ and $f(7) = -2$ , what is the $x$ -coordinate of the vertex of the parabola?

4

2

-2

5

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The correct answer is 2, since the vertex lies on the axis of symmetry, which is the midpoint of the symmetric x-coordinates.

The vertex of a parabola in the

xy

-plane lies on its axis of symmetry. For any quadratic function, points with the same

y

-coordinate are symmetric with respect to the axis of symmetry. Since

f(-3) = -2

and

f(7) = -2

, the

y

-coordinates are equal, which means the axis of symmetry is the vertical line midway between

x = -3

and

x = 7

. The

x

-coordinate of the vertex is the midpoint of these two

x

-values, calculated as

\frac{-3 + 7}{2} = 2

.

Adım Adım Çözüm

1

Identify the two points on the parabola with the same y-value.

The points are

(-3, -2)

and

(7, -2)

.

Since both points share the same y-value of

-2

, they are symmetric with respect to the axis of symmetry of the parabola.

2

Find the x-coordinate of the axis of symmetry by calculating the midpoint of the x-coordinates of the two symmetric points.

\text{Midpoint} = \frac{-3 + 7}{2} = \frac{4}{2} = 2

.

The axis of symmetry is always located exactly halfway between any two symmetric points on the parabola.

3

Determine the x-coordinate of the vertex of the parabola.

The x-coordinate of the vertex is 2.

The vertex of a parabola lies on its axis of symmetry, so its x-coordinate is equal to the x-coordinate of the axis of symmetry.

Anahtar Kavram

Symmetry of quadratic graphs and the axis of symmetry

Alternatif Yöntem

Alternatively, any quadratic function that takes the value

-2

at

x = -3

and

x = 7

can be written in the form

f(x) = a(x + 3)(x - 7) - 2

for some constant

a

. Expanding this expression yields

f(x) = a(x^2 - 4x - 21) - 2

. Since the

x

-coordinate of the vertex of a quadratic function in standard form

y = ax^2 + bx + c

is given by

-\frac{b}{2a}

, we can find the vertex

x

-coordinate as

-\frac{-4a}{2a} = 2

.

Tahmini Süre:1m 30s

Soru 417Soru →

The function $f$ is defined by $f(x) = |x - 6| + 3$ . In the $xy$ -plane, the graph of the function $g$ is obtained by reflecting the graph of $f$ across the $x$ -axis and then translating it vertically upward by 11 units. If $g(a) = 0$ and $a > 0$ , what is the value of $a$ ?

Cevabı ve açıklamayı göster

Cevap: 14

Cevap

The value of

a

is 14.

Reflecting

f(x) = |x - 6| + 3

across the

x

-axis negates the entire expression, giving

-f(x) = -|x - 6| - 3

. Translating this graph upward by 11 units adds 11 to the expression, which yields

g(x) = -|x - 6| - 3 + 11 = -|x - 6| + 8

. Setting

g(a) = 0

results in

-|a - 6| + 8 = 0

, which simplifies to

|a - 6| = 8

. The solutions to this equation are

a = 14

and

a = -2

. Since the problem specifies that

a > 0

, the correct answer is 14.

Adım Adım Çözüm

1

Reflect the function

f(x)

across the

x

-axis.

-f(x) = -|x - 6| - 3

Reflecting a graph across the

x

-axis negates the entire function expression, transforming

y = f(x)

to

y = -f(x)

.

2

Translate the reflected function upward by 11 units to obtain

g(x)

.

g(x) = -|x - 6| + 8

Translating a function vertically upward by

k

units adds

k

to the expression, so

g(x) = -f(x) + 11 = -|x - 6| - 3 + 11 = -|x - 6| + 8

.

3

Set

g(a) = 0

and solve for

a

.

|a - 6| = 8 \implies a = 14

or

a = -2

Solving the equation

-|a - 6| + 8 = 0

requires isolating the absolute value term to get

|a - 6| = 8

. This splits into two cases:

a - 6 = 8

and

a - 6 = -8

.

4

Apply the constraint

a > 0

to identify the final answer.

a = 14

The problem specifies that

a

must be positive, which excludes the solution

a = -2

and leaves

a = 14

.

Anahtar Kavram

Function transformations including reflections across the axes and vertical translations.

Tahmini Süre:1m 30s

Soru 418Soru →

The function $g$ is defined by $g(x) = f(x - 3) - 2$ . If the point $(5, 7)$ lies on the graph of $y = f(x)$ in the $xy$ -plane, which of the following points must lie on the graph of $y = g(x)$ ?

(2, 9)

(8, 5)

(2, 5)

(8, 9)

Cevabı ve açıklamayı göster

Cevap: $(8, 5)$

Cevap

The point

(8, 5)

The transformation

g(x) = f(x - 3) - 2

shifts the graph of

f(x)

to the right by 3 units and down by 2 units. Therefore, the x-coordinate of the point

(5, 7)

increases by 3 (

5 + 3 = 8

), and the y-coordinate decreases by 2 (

7 - 2 = 5

). This determines that the point

(8, 5)

must lie on the graph of

y = g(x)

.

Adım Adım Çözüm

1

Identify the horizontal and vertical transformations applied to the function

f(x)

to get

g(x)

.

The graph of

g(x) = f(x - 3) - 2

is translated horizontally to the right by 3 units and vertically down by 2 units relative to the graph of

f(x)

.

By function transformation rules,

f(x - h) + k

translates the graph horizontally by

h

units (right if

h > 0

) and vertically by

k

units (up if

k > 0

).

2

Calculate the new x-coordinate for the point on the transformed graph.

The new x-coordinate is

5 + 3 = 8

.

Translating the graph to the right by 3 units increases all x-coordinates of points on the graph by 3.

3

Calculate the new y-coordinate for the point on the transformed graph.

The new y-coordinate is

7 - 2 = 5

.

Translating the graph down by 2 units decreases all y-coordinates of points on the graph by 2.

Anahtar Kavram

Function notation and translations in the coordinate plane

Tahmini Süre:1m 15s

Soru 419Soru →

The table below shows several values of $x$ and the corresponding values of the third-degree polynomial function $f$ .

$x$	$f(x)$
$-1$	$0$
$2$	$0$
$5$	$0$
$0$	$30$

If $f(x) = a(x-r_1)(x-r_2)(x-r_3)$ for all real numbers $x$ , where $a$ , $r_1$ , $r_2$ , and $r_3$ are constants, what is the value of $a$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

Since the function

f(x)

is a third-degree polynomial with roots at

x = -1

,

x = 2

, and

x = 5

, it can be factored as

f(x) = a(x+1)(x-2)(x-5)

. Evaluating this expression at

x = 0

gives

f(0) = a(1)(-2)(-5) = 10a

. From the table,

f(0) = 30

, so setting

10a = 30

yields

a = 3

.

Adım Adım Çözüm

1

Identify the roots and factors of the polynomial

f(x)

using the table.

The roots are

x = -1

,

x = 2

, and

x = 5

, which correspond to the factors

(x+1)

,

(x-2)

, and

(x-5)

.

Points where

f(x) = 0

represent the

x

-intercepts (roots) of the function.

2

Write the general form of the cubic polynomial using its factors.

f(x) = a(x+1)(x-2)(x-5)

A third-degree polynomial with three distinct real roots can be factored completely as

a(x-r_1)(x-r_2)(x-r_3)

.

3

Substitute the point

(0, 30)

into the equation to solve for the constant coefficient

a

.

a = 3

The table gives

f(0) = 30

, which allows us to set up the equation

30 = a(0+1)(0-2)(0-5)

and solve for

a

.

Anahtar Kavram

Using the relationship between the factors, roots, and points on the graph of a polynomial function to determine its equation.

Soru 420Soru →

In the $xy$ -plane, the graph of the quadratic function $f$ , defined by $f(x) = -(x - 4)^2 + c$ where $c$ is a constant, intersects the $x$ -axis at two points. If the distance between these two points is $10$ , what is the value of $c$ ?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

25

The quadratic function is defined by

f(x) = -(x - 4)^2 + c

, which is in the vertex form

f(x) = a(x - h)^2 + k

. The vertex of this parabola is at

(4, c)

, and the vertical line

x = 4

is its axis of symmetry. The distance between the two

x

-intercepts is

10

. By symmetry, the intercepts must be located

5

units to the left and right of the axis of symmetry, placing them at

x = 4 - 5 = -1

and

x = 4 + 5 = 9

. Since the graph intersects the

x

-axis at these points, substituting either

x

-coordinate into the function gives a

y

-value of

0

. Substituting

x = 9

yields

0 = -(9 - 4)^2 + c

, which simplifies to

0 = -25 + c

, so

c = 25

.

Adım Adım Çözüm

1

Identify the axis of symmetry of the quadratic function.

The axis of symmetry is

x = 4

.

The function

f(x) = -(x - 4)^2 + c

is written in vertex form,

y = a(x - h)^2 + k

, where the vertex is

(h, k)

and the axis of symmetry is

x = h

.

2

Find the

x

-coordinates of the

x

-intercepts using the distance between them.

The

x

-intercepts are at

x = -1

and

x = 9

.

Since the parabola is symmetric about the line

x = 4

and the distance between the two intercepts is

10

, each intercept is

\frac{10}{2} = 5

units away from the axis of symmetry. Thus, the intercepts are at

4 - 5 = -1

and

4 + 5 = 9

.

3

Substitute one of the

x

-intercepts into the function to solve for the constant

c

.

c = 25

Since

(9, 0)

is on the graph,

f(9) = 0

. Substituting this gives

0 = -(9 - 4)^2 + c \implies 0 = -25 + c \implies c = 25

.

Anahtar Kavram

Using symmetry and the vertex form of a quadratic function to determine constants from key graphical features such as

x

-intercepts.

Advanced Math

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Alternatif Yöntem

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