How do you find the equation of a line tangent to the function #y=x^3-5x# at x=1?

1 Answer
Aug 15, 2016

#y=-2x-2#

Explanation:

Given -

#y=x^3-5x#

Slope of the curve at any point on the curve is given by its
first derivative

#dy/dx=3x^2-5#

Slope of the curve exactly at #x=1#

Substitute #x=1# in the first derivative.

#dy/dx=3(1^2)-5=3-5 = -2#

The slope of the tangent is #m=-2#

The tangent is passing through the point #x=1#

To find the equation of the tangent , we must know the
y-coordinate at point #x=1#

For this substitute #x=1# in the given function

#y=1^3-5(1)=1-5=-4#

Point #(1, -4) # is on the line as well as on the tangent.

We know the slope of the tangent #(m=-2)# and the point
through which it passes #(1, -4)#

#mx+c=y#

#(-2)(1)+c=-4#
#c=-4+2=-2#

Now we have Y- intercept #c=-2# and slope #m=-2#

The equation of the tangent is -

#y=mx+c#

#y=-2x-2#

Look at the graph