A quadratic function f is defined by f(x) = a(x - h)^2 + k, where a, h, and k are constant… — SAT

A quadratic function $f$ is defined by $f(x) = a(x - h)^2 + k$ , where $a$ , $h$ , and $k$ are constants. The graph of $y = f(x)$ in the $xy$ -plane passes through the points $(0, 5)$ and $(4, 5)$ . If the minimum value of $f(x)$ for $0 \le x \le 3$ is $1$ , what is the value of $f(6)$ ?

Answer: 17

Answer

The value of

f(6)

17

The correct answer is

17

. Since the quadratic function passes through

(0, 5)

and

(4, 5)

, its axis of symmetry is the line

x = 2

, which means the vertex

x

-coordinate is

h = 2

. For the interval

0 \le x \le 3

, this vertex is within the bounds. An upward-opening parabola has its minimum value at its vertex, so the minimum value of

1

must be the

y

-coordinate of the vertex, giving

k = 1

. The function can then be written as

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

. Substituting

(0, 5)

into this equation gives

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

, which simplifies to

4a = 4

, meaning

a = 1

. The fully determined function is

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

. Evaluating this at

x = 6

yields

f(6) = (6 - 2)^2 + 1 = 17

Step-by-Step Solution

Determine the axis of symmetry of the parabola.

h = 2

Since the points

(0, 5)

and

(4, 5)

have the same

y

-value, they are symmetric across the vertical line of symmetry of the parabola. The

x

-coordinate of the vertex

h

is the average of the

x

-coordinates of these two points:

h = \frac{0 + 4}{2} = 2

Determine the

y

-coordinate of the vertex

k

using the minimum value over the interval

[0, 3]

k = 1

The vertex

x

-coordinate

x = 2

lies within the interval

0 \le x \le 3

. For a quadratic function that opens upward, the minimum value occurs at the vertex. Thus, the minimum value on this interval is the vertex

y

-coordinate, so

k = 1

Substitute the point

(0, 5)

into the vertex form of the equation to solve for

a

a = 1

Using the vertex form

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

, substitute

x = 0

and

f(x) = 5

5 = a(0 - 2)^2 + 1 \implies 5 = 4a + 1 \implies 4a = 4 \implies a = 1

Evaluate

f(6)

using the complete quadratic function formula.

f(6) = 17

Substitute

x = 6

into the equation

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

f(6) = (6 - 2)^2 + 1 = 4^2 + 1 = 16 + 1 = 17

Key Concept

Using symmetry properties and interval extrema to determine the equation of a quadratic function in vertex form.

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