## Calculus 10th Edition

(a) $(y-4)=-2(x+1)$
(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 $(y-y_{1})=m(x-x_{1})$ $(y-4)=-2(x+1)$ This is the point-slope form.