The graph of the quadratic function f(x) = ax^2 + bx + c, where a, b, and c are constants,… — SAT

The graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, is a parabola in the $xy$ -plane that passes through the points $(-3, 22)$ and $(9, 22)$ . If the minimum value of $f(x)$ is $4$ , what is the value of $f(1)$ ?

Answer: 6

Answer

The correct answer is 6.

The correct answer is 6. The axis of symmetry of the parabola is halfway between the points with equivalent y-values: x = 3. Using the minimum value of 4, the vertex is identified as (3, 4). Writing the equation in vertex form as f(x) = a(x - 3)^2 + 4 and substituting (9, 22) yields a = 0.5. Evaluating the function f(x) = 0.5(x - 3)^2 + 4 at x = 1 yields 6.

Step-by-Step Solution

Determine the axis of symmetry of the parabola.

The axis of symmetry is

x = 3

Because the parabola passes through the points

(-3, 22)

and

(9, 22)

, which have the same

y

-coordinate, the axis of symmetry must lie halfway between their

x

-coordinates:

x = \frac{-3 + 9}{2} = 3

Write the quadratic function in vertex form.

f(x) = a(x - 3)^2 + 4

Since the function has a minimum value of

4

, the vertex of the upward-opening parabola is at

(3, 4)

Solve for the leading coefficient

a

a = 0.5

Substitute the point

(9, 22)

into the vertex form equation:

22 = a(9 - 3)^2 + 4

, which simplifies to

18 = 36a

, so

a = 0.5

Evaluate

f(1)

f(1) = 6

Substitute

x = 1

into the completed function

f(x) = 0.5(x - 3)^2 + 4

to get

f(1) = 0.5(1 - 3)^2 + 4 = 0.5(4) + 4 = 6

Key Concept

Using symmetry, vertex form, and given points to determine a quadratic function's equation and evaluate it.

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