# Calc tangent problem

1. Apr 20, 2009

### coookiemonste

1. The problem statement, all variables and given/known data
Let f(x)=a(7-x^2)
find in terms of a, the equations of the lines tangent to these curves at x=-1.

2. Relevant equations
?

3. 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 dont know where I went wrong.

2. Apr 20, 2009

### Cyosis

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

3. Apr 20, 2009

### 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?

4. Apr 20, 2009

### coookiemonste

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) ?

5. Apr 20, 2009

### dx

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

6. Apr 20, 2009

### coookiemonste

o ok.
but wouldnt you still get y-8a=2a(x+1)
which is y=2ax+10a?

7. Apr 20, 2009

### dx

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

8. Apr 25, 2009

### coookiemonste

thanks!

9. Apr 25, 2009

### dx

No problem.

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