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