Find the equation

Question:
Find the equation of the tangent line to the curve $y=x^{2}+4 x-16$ which is parallel to the line $3 x-y+1=0$.

Solution:

finding the slope of the tangent by differentiating the curve

$\frac{d y}{d x}=2 x+4$

$m($ tangent $)=2 x+4$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

now comparing the slope of a tangent with the given equation

$2 x+4=3$

$x=-\frac{1}{2}$

Now substituting the value of $x$ in the curve to find $y$

$y=\frac{1}{4}-2-16=-\frac{71}{4}$

Therefore, the equation of tangent parallel to the given line is

$y+\frac{71}{4}=3\left(x+\frac{1}{2}\right)$