In the xy-plane, the graph of the quadratic function f(x) = -(x - d)^2 + d^2, where d is a… — SAT

In the $xy$ -plane, the graph of the quadratic function $f(x) = -(x - d)^2 + d^2$ , where $d$ is a positive constant, has vertex $V$ and intersects the $x$ -axis at points $P$ and $Q$ . If the area of triangle $PVQ$ is $64$ , what is the value of $d$ ?

A
2
4Cevap
C
8
D
16

Cevap

The value of

d

4

The function

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

is in vertex form,

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

, so its vertex is

V(d, d^2)

. Since

d > 0

, the vertex is in the first quadrant, and the height of the triangle is

d^2

. Setting

f(x) = 0

gives the

x

-intercepts

P(0,0)

and

Q(2d,0)

, meaning the base of the triangle has a length of

2d

. Using the area of a triangle formula, the area is

\frac{1}{2} \times 2d \times d^2 = d^3

. Since the area is given as

64

, we set

d^3 = 64

, which yields

d = 4

Adım Adım Çözüm

Identify the vertex

V

of the quadratic function

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

The vertex is

V(d, d^2)

The function is written in vertex form,

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

, where the vertex is

(h, k)

Find the

x

-intercepts

P

and

Q

of the function by setting

f(x) = 0

x = 0

and

x = 2d

Setting

-(x - d)^2 + d^2 = 0

gives

(x - d)^2 = d^2

, which simplifies to

x - d = \pm d

Calculate the area of triangle

PVQ

in terms of

d

The area is

d^3

The base of the triangle along the

x

-axis is the distance between the intercepts,

2d - 0 = 2d

. The height is the

y

-coordinate of the vertex,

d^2

. The area is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (2d)(d^2) = d^3

Solve for

d

using the given area of

64

d = 4

Setting the area expression

d^3

equal to

64

and taking the cube root of both sides gives

d = 4

Anahtar Kavram

Finding the vertex and intercepts of a quadratic function in vertex form and applying geometric formulas to analyze the graph.

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