Question

How do you find the equation of the line tangent to the graph of `f(x)=2x^2` at x=-1?

Answer 1

y = 4x + 1

Explanation:

Remember this-

To find the equation of a straight line, we need two information –

They are (i) slope of the line and (ii) (x, y) co-ordinates of a point on that line.

Tangent is a straight line.

The given function is
y = `2x^2`
It is a U shaped curve or parabola. It doesn’t have uniform slope throughout its length.

Its slope at any given point is its first derivative.

`dy/dx`= 4x

At x = 1, its slope is dy/dx = 4(1) = 4

At x = 1 , the slope of the curve is 4.

A tangent is drawn to that point. To find the y co-ordinate of the point substitute x = 1 in the given function.

Y = `2 xx 1^2` = 2

(1, 2) is a point on the tangent. The slope of the tangent is m = 4.
The equation of the tangent is
y – y1 = m(x – x1)
y – (2) = 4(x – 1)
y - 2 = 4x – 1
y = 4x – 1 + 2

y = 4x + 1