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Question
if $f(x)=x^2 + 4x - 4$, simplify each of the following.
$f(x + h) = \square$
$f(x + h) - f(x) = \square$
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Part 1: Find \( f(x + h) \)
Step 1: Substitute \( x + h \) into \( f(x) \)
We have \( f(x)=x^{2}+4x - 4 \). Replace \( x \) with \( x + h \):
\( f(x + h)=(x + h)^{2}+4(x + h)-4 \)
Step 2: Expand \( (x + h)^{2} \) and \( 4(x + h) \)
Using the formula \( (a + b)^{2}=a^{2}+2ab + b^{2} \), \( (x + h)^{2}=x^{2}+2xh+h^{2} \). And \( 4(x + h)=4x + 4h \). So:
\( f(x + h)=x^{2}+2xh+h^{2}+4x + 4h-4 \)
Part 2: Find \( f(x + h)-f(x) \)
Step 1: Substitute \( f(x + h) \) and \( f(x) \)
We know \( f(x + h)=x^{2}+2xh+h^{2}+4x + 4h-4 \) and \( f(x)=x^{2}+4x - 4 \). Then:
\( f(x + h)-f(x)=(x^{2}+2xh+h^{2}+4x + 4h-4)-(x^{2}+4x - 4) \)
Step 2: Distribute the negative sign and simplify
Distribute the negative sign to each term in \( f(x) \):
\( f(x + h)-f(x)=x^{2}+2xh+h^{2}+4x + 4h-4 - x^{2}-4x + 4 \)
Now, combine like terms:
- The \( x^{2} \) terms: \( x^{2}-x^{2}=0 \)
- The \( 4x \) terms: \( 4x-4x = 0 \)
- The constant terms: \( - 4 + 4=0 \)
We are left with \( 2xh+h^{2}+4h \)
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For \( f(x + h) \): \( \boldsymbol{x^{2}+2xh + h^{2}+4x + 4h-4} \)
For \( f(x + h)-f(x) \): \( \boldsymbol{2xh+h^{2}+4h} \)