Answer
$a.\quad y=-x+3$
$b.\displaystyle \quad c=\frac{1}{2}$
$c.\quad y=-x+5.25$
$ d.\quad$ See image
Work Step by Step
(a) Slope
$m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\displaystyle \frac{1-4}{2+1}=-1$
Secant line:$\qquad y-y_{1}=m(x-x_{1})$
$y-4=-(x+1)$
$y=-x+3$
(b)
If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ ,
then there exists a number $c$ in $(a, b)$ such that$ f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}.$
$f^{\prime}(x)=-2x$
$f^{\prime}(x)=m$
$-2x=-1$
$x=\displaystyle \frac{1}{2} \qquad(c=\frac{1}{2})$
(c)
$f(c)=f(\displaystyle \frac{1}{2})=-\frac{1}{4}+5=\frac{19}{4}$
Tangent point: $(\displaystyle \frac{1}{2},\frac{19}{4})$
Tangent line: $\qquad y-y_{1}=m(x-x_{1})$
$y-\displaystyle \frac{19}{4}=-(x-\frac{1}{2})$
$4y-19=-4x+2$
$4y=-4x+21$
$y=-x+5.25$
