High School

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Evaluate the function [tex]f(x) = x^2 - 2x + 7[/tex] at the given values of the independent variable and simplify.

a. [tex]f(-2)[/tex]
b. [tex]f(x+7)[/tex]
c. [tex]f(-x)[/tex]

a. [tex]f(-2) = 15[/tex] (Simplify your answer.)
b. [tex]f(x+7) = x^2 + 12x + 42[/tex] (Simplify your answer.)
c. [tex]f(-x) = \square[/tex] (Simplify your answer.)

Answer :

Sure! Let's evaluate the function [tex]\( f(x) = x^2 - 2x + 7 \)[/tex] at different values.

### a. Evaluate [tex]\( f(-2) \)[/tex]

1. Substitute [tex]\(-2\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[
f(-2) = (-2)^2 - 2(-2) + 7
\][/tex]

2. Calculate [tex]\((-2)^2\)[/tex]:
[tex]\[
(-2)^2 = 4
\][/tex]

3. Calculate [tex]\(-2 \times (-2)\)[/tex]:
[tex]\[
-2 \times (-2) = 4
\][/tex]

4. Substitute these values back into the equation:
[tex]\[
f(-2) = 4 + 4 + 7
\][/tex]

5. Add the numbers:
[tex]\[
f(-2) = 15
\][/tex]

So, [tex]\( f(-2) = 15 \)[/tex].

### b. Evaluate [tex]\( f(x + 7) \)[/tex]

1. Substitute [tex]\( (x + 7) \)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[
f(x + 7) = (x + 7)^2 - 2(x + 7) + 7
\][/tex]

2. Expand [tex]\((x + 7)^2\)[/tex]:
[tex]\[
(x + 7)^2 = x^2 + 14x + 49
\][/tex]

3. Distribute the [tex]\(-2\)[/tex] across [tex]\((x + 7)\)[/tex]:
[tex]\[
-2(x + 7) = -2x - 14
\][/tex]

4. Substitute these into the equation:
[tex]\[
f(x + 7) = x^2 + 14x + 49 - 2x - 14 + 7
\][/tex]

5. Combine like terms:
[tex]\[
f(x + 7) = x^2 + (14x - 2x) + (49 - 14 + 7) = x^2 + 12x + 42
\][/tex]

So, [tex]\( f(x + 7) = x^2 + 12x + 42 \)[/tex].

### c. Evaluate [tex]\( f(-x) \)[/tex]

1. Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[
f(-x) = (-x)^2 - 2(-x) + 7
\][/tex]

2. Calculate [tex]\((-x)^2\)[/tex]:
[tex]\[
(-x)^2 = x^2
\][/tex]

3. Calculate [tex]\(-2 \times (-x)\)[/tex]:
[tex]\[
-2 \times (-x) = 2x
\][/tex]

4. Substitute these values back into the equation:
[tex]\[
f(-x) = x^2 + 2x + 7
\][/tex]

So, [tex]\( f(-x) = x^2 + 2x + 7 \)[/tex].

These are the simplified answers for each part of the function evaluation!

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