# Calc Tangent Problem: Find Lines Tangent to f(x) at x=-1

• coookiemonste
In summary, the conversation was about finding the equations of the tangent lines to the curve f(x)=a(7-x^2) at x=-1. The solution involved taking the derivative, which was found to be -2ax, and then using the point-slope form to get the equation y=2ax+8a. However, there was a mistake in calculating the y-intercept, which should have been 6a instead of 8a.

## Homework Statement

Let f(x)=a(7-x^2)
find in terms of a, the equations of the lines tangent to these curves at x=-1.

?

## The Attempt at a Solution

So I took the derivative of f(x).
f'(x)=a(-2x)+7-x^2
then i plugged in -1 into f'(x)
and i got f'(-1)=2ax+7+1=2ax+8
but the answer to the problem is y=2ax+8a.
I don't know where I went wrong.

It went wrong at the part where you took the derivative. Write out the brackets and then ask yourself what's $$\frac{d7a}{dx}, \frac{-dx^2}{dx}$$.

The derivative of a(7 - x²) is -2ax.

The slope of a line is not enough to determine the equation of the line. If you have the slope, and also its height at a point b, call it c, then the equation of the line will be:

y = c + (x - b)(slope)

Do you recognize this?

ok
so f'(x)=-2ax which is the slope.
then to get the set of points
you know x=-1,
so f(-1)=8a.
but if you used point slope format
wouldnt you get y-8a=-2ax(x+1) ?

Yes, except you have to replace -2ax in your equation with -2a(-1) since you want the slope at -1.

o ok.
but wouldn't you still get y-8a=2a(x+1)
which is y=2ax+10a?

You calculated f(-1) wrong. It should be 6a not 8a.

thanks!

No problem.