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

Question

Luke Alvarez · Accepted Answer

y = 4x + 1

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

Aurora Alvarado · Answer

undefined To find the equation of the line tangent to the graph of f(x)=2x^2 at x=-1, we need to find the slope of the tangent line and the point of tangency.

First, we find the derivative of f(x) with respect to x, which gives us f'(x)=4x.

Next, we substitute x=-1 into f'(x) to find the slope of the tangent line at x=-1. So, f'(-1)=4(-1)=-4.

Now, we have the slope of the tangent line, which is -4.

To find the point of tangency, we substitute x=-1 into f(x). So, f(-1)=2(-1)^2=2.

Therefore, the point of tangency is (-1, 2).

Finally, we can use the point-slope form of a linear equation, y-y1=m(x-x1), where m is the slope and (x1, y1) is a point on the line.

Substituting the values, we get the equation of the tangent line as y-2=-4(x+1).

Simplifying this equation gives us the final answer: y=-4x-2.