What is the equation of the line tangent to #f(x)=9x^2 - 28x - 34 # at #x=-1#?

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Answer 1 · 2016-03-23T14:47:12

#y=-64x+67#

#f(x)=9x^2-28x-34#

The tangent line is of the form #y=mx+c#. We can get #m# by taking the 1st derivative:

#f'(x)=18x-28#

So when #x=-1# then #m=(18xx-1)-28=-64#

To find the #y# coordinate:

#f(-1)=9+28-34=3#

Putting these values into the tangent equation to get #crArr#

#3=(-64xx-1)+c#

#:.c=67#

So the equation of the tangent line #rArr#

#y=-64x+67#

graph{(-64x+67-y)(9x^2-28x-34-y)=0 [-160, 160, -80, 80]}