How do you write the equation for the line tangent to #f(x)# in #x = 0# if f is the function given by #f(x) = (2x^2 + 5x -1)^7#?

1 Answer
Gió
Mar 12, 2015

First evaluate the derivative of your function and substitute #x=0# in it:
#f'(x)=7(2x^2+5x-1)^6*(4x+5)# where I used the Chain Rule;
substitute #x=0#
#f'(0)=7*5=35# which is the SLOPE of your line.

Now set #x=0# in your function to find the point where the line is tangent to your curve (represented by #f(x)#):
#f(0)=-1#.

So you need a line with slope #35# and passing through (#0,-1#):
You can use the general form for the line of slope #m# passing through (#x_0,y_0#) as:
#y-y_0=m(x-x_0)#
#y-(-1)=35(x-0)#
#y=35x-1#

As an illustration you have:
enter image source here

hope it helps

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