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\[f(x)=x^2+5x+1 \]put into the differance quo =2x+5+h
yes. \[\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2+5(x+h)+1-(x^2+5x+1)}{h}\] \[\frac{x^2+2xh+h^2+5x+5h+1-x^2-5x-1}{h}\] \[\frac{2xh+5h+h^2}{h}\] \[2x+5+h\]
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