In the xy-plane, the graph of the quadratic function f(x) = ax^2 + bx + c, where a, b, and… — SAT

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, has a vertex $(h, k)$ that lies on the line $y = 4x + 8$ . If the graph of $y = f(x)$ has $x$ -intercepts at $x = -2$ and $x = 6$ , what is the value of $c$ ?

Cevap: 12

Cevap

The value of c is 12.

The correct answer is 12. The axis of symmetry of the quadratic function lies midway between its x-intercepts at

x = -2

and

x = 6

, giving an x-coordinate of

h = 2

for the vertex. Since the vertex lies on the line

y = 4x + 8

, its y-coordinate is

k = 4(2) + 8 = 16

. Substituting the vertex

(2, 16)

into the vertex form of a quadratic function gives

f(x) = a(x - 2)^2 + 16

. Using the x-intercept at

(6, 0)

to solve for

a

gives

0 = a(6 - 2)^2 + 16

, which yields

a = -1

. Evaluating the function at

x = 0

to find the constant term

c

gives

c = f(0) = -(0 - 2)^2 + 16 = 12

Adım Adım Çözüm

Determine the x-coordinate of the vertex of the quadratic function.

h = 2

The axis of symmetry of a parabola lies midway between its x-intercepts. The midpoint of the intercepts

x = -2

and

x = 6

is calculated as

\frac{-2 + 6}{2} = 2

Determine the y-coordinate of the vertex using the given linear equation.

k = 16

Since the vertex

(h, k)

lies on the line

y = 4x + 8

, substituting

x = h = 2

into the line equation gives

k = 4(2) + 8 = 16

Write the quadratic function in vertex form and solve for the leading coefficient

a

a = -1

The vertex form is

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

. Substituting the vertex

(2, 16)

yields

f(x) = a(x - 2)^2 + 16

. Using the x-intercept

(6, 0)

, substitute

x = 6

and

f(x) = 0

into the equation:

0 = a(6 - 2)^2 + 16

, which simplifies to

16a = -16

, resulting in

a = -1

Expand the quadratic function or evaluate it at

x = 0

to find the value of

c

c = 12

In standard form

f(x) = ax^2 + bx + c

, the constant term

c

is equal to

f(0)

. Substituting

x = 0

into

f(x) = -(x - 2)^2 + 16

yields

f(0) = -(0 - 2)^2 + 16 = -4 + 16 = 12

Anahtar Kavram

Using symmetry and the vertex form of a quadratic function to determine its standard form coefficients.

Alternatif Yöntem

Alternatively, since the x-intercepts are

-2

and

6

, the quadratic function can be written in factored form as

f(x) = a(x + 2)(x - 6)

. Expanding this gives

f(x) = a(x^2 - 4x - 12) = ax^2 - 4ax - 12a

. Comparing this to the standard form

f(x) = ax^2 + bx + c

, we see that the x-coordinate of the vertex is

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

. Using the line equation

y = 4x + 8

x = 2

, we find the vertex y-coordinate is

k = 16

. Since the vertex is

(2, 16)

, we evaluate the factored form at

x = 2

16 = a(2 + 2)(2 - 6) \Rightarrow 16 = -16a \Rightarrow a = -1

. The constant term

c

-12a

, so

c = -12(-1) = 12

Tahmini Süre:2m 30s

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