Find the equation of the tangent line to the curve y = x2 + 4x – 16

1.	Find the equation of the tangent line to the curve y = x2 + 4x – 16 which is parallel to the line 3x – y + 1 = 0.
Answer» finding the slope of the tangent by differentiating the curve \(\frac{dy}{dx}=2x+4\) m(tangent) = 2x + 4 equation of tangent is given by y – y₁ = m(tangent)(x – x₁) now comparing the slope of a tangent with the given equation 2x + 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 \(y+\frac{71}{4}=3(x+\frac{1}{2})\)

Find the equation of the tangent line to the curve y = x2 + 4x – 16 which is parallel to the line 3x – y + 1 = 0.

Answer»

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=2x+4\)

m(tangent) = 2x + 4

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

2x + 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

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