Question

for the function $f(x) = 4x^2 - 4x + 6$, evaluate and fully simplify each of the following.
$f(x + h) = \square$
$\frac{f(x + h) - f(x)}{h} = \square$
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Explanation:

Part 1: Find \( f(x + h) \)

Step 1: Substitute \( x + h \) into \( f(x) \)

We have \( f(x)=4x^{2}-4x + 6 \). Replace \( x \) with \( x + h \):
\( f(x + h)=4(x + h)^{2}-4(x + h)+6 \)

Step 2: Expand \( (x + h)^{2} \)

Using the formula \( (a + b)^{2}=a^{2}+2ab + b^{2} \), where \( a = x \) and \( b = h \), we get \( (x + h)^{2}=x^{2}+2xh+h^{2} \). So:
\( f(x + h)=4(x^{2}+2xh + h^{2})-4(x + h)+6 \)

Step 3: Distribute the constants

Distribute 4 in \( 4(x^{2}+2xh + h^{2}) \) and - 4 in \( -4(x + h) \):
\( f(x + h)=4x^{2}+8xh+4h^{2}-4x-4h + 6 \)

Part 2: Find \( \frac{f(x + h)-f(x)}{h} \)

Step 1: Substitute \( f(x + h) \) and \( f(x) \)

We know \( f(x + h)=4x^{2}+8xh + 4h^{2}-4x-4h + 6 \) and \( f(x)=4x^{2}-4x + 6 \). So:
\( \frac{f(x + h)-f(x)}{h}=\frac{(4x^{2}+8xh+4h^{2}-4x - 4h + 6)-(4x^{2}-4x + 6)}{h} \)

Step 2: Simplify the numerator

Remove the parentheses and combine like terms:
\( (4x^{2}+8xh+4h^{2}-4x - 4h + 6-4x^{2}+4x - 6)=8xh+4h^{2}-4h \)

Step 3: Factor out \( h \) from the numerator

\( 8xh + 4h^{2}-4h=h(8x + 4h-4) \)

Step 4: Cancel out \( h \) (assuming \( h

eq0 \))
\( \frac{h(8x + 4h-4)}{h}=8x + 4h-4 \)

Answer:

\( f(x + h)=4x^{2}+8xh + 4h^{2}-4x-4h + 6 \)
\( \frac{f(x + h)-f(x)}{h}=8x + 4h-4 \)