How do you find the equation of the tangent and normal line to the curve y=x3 at x=2?

Question

Mey9ci0 · Accepted Answer

The slope of the tangent is given by value of first derivative at that point i.e. here at x=2
As

y = x^{3}

, we have

\frac{d y}{d x} = 3 x^{2}

and at x=2, its value is

3 \cdot 2^{2} = 12

When we seek a tangent at x=2, it means tangent at

(2, 2^{3})

i.e. at (2, 8)
As tangent passes through (2,8) and has a slope 12, its equation is
y-8=12(x-2) or 12x-y-16=0
As normal is perpendicular to tangent, its slope is

\frac{- 1}{12}

and hence equation of normal is

y - 8 = - \frac{1}{12} (x - 2)

or x+12y-98=0

Ausspruchx807 · Accepted Answer

Given: f(x)=x3 and x0=2Find the value of the function at the given point: y0=f(2)=8The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function,evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(x3)′=3x2Hence, M(x0)=−1f′(x0)=−13x02Next, find the slope at the given point.m=M(2)=−112Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−8=−x−212Or, more simply: y=496−x12.

How do you find the equation of the

Answered question

Answer & Explanation

New Questions in Calculus 1