Question

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`?

Answer 1

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