The quadratic function f is defined by f(x) = ax^2 + bx + c, where a, b, and c are constan… — SAT

The quadratic function $f$ is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, and has its vertex at $(2, 4)$ . The function $g$ is defined by $g(x) = f(x - 3) + k$ , where $k$ is a constant. The vertex of the graph of $y = g(x)$ in the $xy$ -plane lies on the line $y = 2x - 3$ . If the product of the $x$ -intercepts of the graph of $y = g(x)$ is $11$ , what is the $y$ -intercept of the graph of $y = f(x)$ ?

2Cevap
B
6
C
-4
D
-6

Cevap

The correct answer is 2. By writing the quadratic function

f(x)

in vertex form as

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

, we can apply the transformation rules to express

g(x) = f(x - 3) + k

as having a vertex at

(5, 4 + k)

. Since this vertex lies on the line

y = 2x - 3

, substituting

x = 5

gives

4 + k = 7

, meaning

g(x) = a(x - 5)^2 + 7

. The roots of

g(x) = 0

are

5 \pm \sqrt{-7/a}

, and their product is

25 + 7/a

. Setting this product equal to 11 yields

a = -1/2

. Substituting

a = -1/2

back into the expression for

f(0) = 4a + 4

gives 2.

Adım Adım Çözüm

Write the function

f(x)

in vertex form using its vertex

(2, 4)

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

Any quadratic function with a vertex at

(h, k)

can be written in the form

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

Determine the vertex of the transformed function

g(x) = f(x - 3) + k

The vertex of

g(x)

(5, 4 + k)

The transformation

f(x - 3)

shifts the graph 3 units to the right, changing the

x

-coordinate of the vertex from 2 to 5. The addition of

k

shifts the graph vertically by

k

units, changing the

y

-coordinate of the vertex from 4 to

4 + k

Use the condition that the vertex of

y = g(x)

lies on the line

y = 2x - 3

to find the value of

k

and the vertex coordinates.

4 + k = 7 \implies k = 3

. The vertex of

g(x)

(5, 7)

Since the vertex

(5, 4 + k)

lies on the line

y = 2x - 3

, substituting

x = 5

into the line equation gives

y = 2(5) - 3 = 7

Write

g(x)

in vertex form and express its roots (the

x

-intercepts) in terms of

a

g(x) = a(x - 5)^2 + 7

. The roots are

x = 5 \pm \sqrt{-\frac{7}{a}}

Setting

g(x) = 0

gives

a(x - 5)^2 + 7 = 0

, which simplifies to

(x - 5)^2 = -\frac{7}{a}

, yielding

x = 5 \pm \sqrt{-\frac{7}{a}}

Calculate the product of the roots and set it equal to 11 to solve for

a

25 + \frac{7}{a} = 11 \implies a = -\frac{1}{2}

The product of the roots is

(5 + \sqrt{-\frac{7}{a}})(5 - \sqrt{-\frac{7}{a}}) = 25 - (-\frac{7}{a}) = 25 + \frac{7}{a}

. Setting this equal to 11 gives

\frac{7}{a} = -14

, so

a = -\frac{1}{2}

Evaluate

f(0)

to find the

y

-intercept of the graph of

f(x)

f(0) = 4(-\frac{1}{2}) + 4 = 2

The

y

-intercept is the value of the function at

x = 0

. Substituting

x = 0

and

a = -\frac{1}{2}

into

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

yields

4a + 4 = 2

Anahtar Kavram

Quadratic transformations and properties of roots in vertex form

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