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

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Answer 1 · 2018-04-14T14:59:36

#y=-5x-8#

We are given: #f(x)=3x^2+x-5#

At #x=-1,f(x)=3*(-1)^2-1-5#

#=3*1-1-5#

#=3-1-5#

#=2-5#

#=-3#

So, the point is at #(-1,-3)#.

We first find the slope of the tangent line by taking the derivative of the function, and plugging in #x=-1# there.

So, #f'(x)=6x+1#

#f'(-1)=6*-1+1#

#=-6+1#

#=-5#

Therefore, the slope of the tangent line will be #-5#.

Now, we use the point-slope form to find the equation of the line, it is:

#y-y_0=m(x-x_0)#

where #(x_0,y_0)# are the original coordinates.

So,

#y-(-3)=-5(x-(-1))#

#y+3=-5(x+1)#

#y+3=-5x-5#

#y=-5x-5-3#

#y=-5x-8#

A graph proves it:

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