What is the equation of the line tangent to #f(x)= 5x^3+4x^2 # at #x=-1#?

Question

5945 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-16T13:55:55

#y = 7x + 6#

graph{5x^3+4x^2 [-10, 10, -5, 5]}

The tangent line passes through #(-1,f(-1))#.

#f(-1) = 5(-1)^3 + 4(-1)^2#

#= -1#

The gradient of the tangent line is #f'(-1)#. To find #f'(-1)#, find #f'(x)# first and substitute #x=-1#.

#f'(x) = 15x^2 + 8x#

#f'(-1) = 15(-1)^2 + 8(-1)#

#= 7#

Know we know the tangent line passes through #(-1,-1)# and has gradient #7#, time to find its equation.

The #y#-intercept is given by

#(-1)-7(-1) = 6#

Equation of tangent line:

#y = 7x + 6#

graph{7x+6 [-10, 10, -5, 5]}