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2 years ago

Please help me, im struggling

Mathematics

1 answer:

Molodets [167]2 years ago

7 0

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

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

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