## Calculus (3rd Edition)

The slope of the secant line=$\frac{f(2+h)-f(2)}{(2+h)-2}=\frac{[2(2+h)^{2}-3(2+h)-5]-[2(2)^{2}-3(2)-5]}{h}$ $=\frac{[2\times2^{2}+2\times h^{2}+2\times4h-6-3h-5]-(-3)}{h}=\frac{2h^{2}+5h}{h}=2h+5$ (a) 7 (b) 5
The slope of the secant line=$\frac{f(2+h)-f(2)}{(2+h)-2}=\frac{[2(2+h)^{2}-3(2+h)-5]-[2(2)^{2}-3(2)-5]}{h}$ $=\frac{[2\times2^{2}+2\times h^{2}+2\times4h-6-3h-5]-(-3)}{h}=\frac{2h^{2}+5h}{h}=2h+5$ (a) Knowing that $h=3-2=1$, we have the slope of the given secant line=$2h+5=2\times1+5=7$ (b) The slope of the tangent line at x=2 is given by $m=\lim\limits_{h \to 0}\frac{f(2+h)-f(2)}{h}$ We found $\frac{f(2+h)-f(2)}{h}$ to be equal to 2h+5. Therefore, $m=\lim\limits_{h \to 0}2h+5=5$