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

`3 x-y+1=0`

`frac{d}{d x}(3 x)-frac{d y}{d x}+frac{d}{d x}(1)=0`

`frac{d y}{d x}=3`

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

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

`text { Hence }`

`2 x+4=3`

`x=frac{3-4}{2}=frac{-1}{2}`

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