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

The quadratic function $f$ is defined by $f(x) = -2x^2 + bx + c$ , where $b$ and $c$ are constants. In the $xy$ -plane, the vertex of the graph of $f$ has an $x$ -coordinate of $3$ . If $f(1) = 2$ , what is the $y$ -coordinate of the vertex of the graph of $f$ ?

Cevap: 10

Cevap

The quadratic function in standard form is

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

, which has a leading coefficient of

a = -2

. The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex of the parabola. Given that the

x

-coordinate of the vertex is

3

(so

h = 3

), we can write the function as

f(x) = -2(x-3)^2 + k

. Since the graph passes through the point

(1, 2)

, we substitute

x = 1

and

f(1) = 2

into the equation:

2 = -2(1-3)^2 + k

. Simplifying the expression gives

2 = -2(-2)^2 + k

, which becomes

2 = -2(4) + k

, or

2 = -8 + k

. Adding

8

to both sides yields

k = 10

. Thus, the

y

-coordinate of the vertex is

10

Adım Adım Çözüm

Write the quadratic function in vertex form.

f(x) = -2(x-3)^2 + k

The standard form equation

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

shows that the leading coefficient

a = -2

. The vertex form is

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

, where

(h, k)

is the vertex. Since the

x

-coordinate of the vertex is

3

, we substitute

a = -2

and

h = 3

into the vertex form.

Substitute the point

(1, 2)

into the vertex form equation.

2 = -2(1-3)^2 + k

Since

f(1) = 2

, the graph passes through the point

(1, 2)

. Substituting these values allows us to solve for the unknown vertex

y

-coordinate,

k

Simplify the equation and solve for

k

k = 10

Simplifying the term

-2(1-3)^2

yields

-2(-2)^2 = -2(4) = -8

. This simplifies the equation to

2 = -8 + k

. Adding

8

to both sides gives

k = 10

Anahtar Kavram

Vertex form of a quadratic function

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru