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

Explanation

Let `(x_{0}, y_{0})` be the point of intersection of both the curve and the tangent.

`y=x^{2}+4 x-16`

Since,`(x_{0}, y_{0})` lies on curve . Therefore

`y_{0}=x_{0}^{2}+4 x_{0}-16 ldots(1)` Now, `y=x^{2}+4 x-16`

`Rightarrow frac{d y}{d x}=2 x+4`

Slope of tangent `=(frac{d y}{d x})_{(x_{0}, y_{0})}=2 x_{0}+4`

Given that The tangent is parallel to the line So,

Slope of tangent=slope of the given line

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