Answer
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
Work Step by Step
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, $m=\lim\limits_{h \to 0}2h+5=5$
