Answer
y=-2x+2
Work Step by Step
(a) Take the derivative of the function $f(x)=x^2+3$:
$f'(x)=2x$
Plug the $x$-value of the given point into the derivative to get the slope of the tangent line:
$f'(-1)=2(-1)=-2$
Plug the slope $m$ and the given point $(x_{1},y_{1})$ into the point-slope formula $(y-y_{1})=m(x-x_{1})$
$(y-4)=-2(x+1)$
y=-2x-2+4
y=-2x+2
This is the point-slope form.
