Answer
$a.$ $f'(x)=2ax+b$
$b.$ $f'(x)=8x-3$
$c.$ $f'(1)=5.$
Work Step by Step
$a.$ By the definition of the derivative we have
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}= \lim_{h\to0}\frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h}=\lim_{h\to0}\frac{a(x^2+h^2+2hx)+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{ax^2+ah^2+2ahx+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{ah^2+2ahx+bh}{h}=\lim_{h\to0}(ah+2ax+b)=a\cdot0+2ax+b=2ax+b.$$
$b.$ We have shown that the derivative of any function of the form of $ax^2+bx+c$ is equal to $2ax+b$ at every point $x$. If we apply that to the given function we get
$$f'(x)=2\cdot4x-3=8x-3.$$
$c.$ We just have to put $x=1$ into the previous expression
$$f'(1) = 8\cdot1-3=5.$$
