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#?

Question

Ezekiel Alexander · Accepted Answer

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:

hope it helps

Emma Aldridge · Answer

undefined To find the equation for the line tangent to f(x) at x = 0, we need to find the derivative of f(x) and evaluate it at x = 0. The derivative of f(x) can be found using the chain rule. The derivative of (2x^2 + 5x - 1)^7 is 7(2x^2 + 5x - 1)^6 * (4x + 5). Evaluating this derivative at x = 0 gives us 7(-1)^6 * 5 = 35. Therefore, the equation for the line tangent to f(x) at x = 0 is y = 35x.