Please help me, im struggling

asked May 7, 2023 141k views

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2 votes

Part 1

f(a)=4-7a+16a^2

Part 2

$f(a+h)=4-7(a+h)+16(a+h)^2\\\\=4-7(a+h)+16(a^2 +2ah+h^2)\\\\=4-7a-7h+16a^2 +32ah+16h^2$

Part 3

$(f(a+h)-f(a))/(h)=((4-7a-7h+16a^2 +32ah+16h^2-(4-7a+16a^2))/(h)\\\\=(-7h+32ah+16h^2)/(h)\\\\=32a+16h-7$

Aviggiano answered May 13, 2023

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