# Math Help - help with a slope of the tangent to a curve

1. ## help with a slope of the tangent to a curve

Here's the problem -

400-6x^2

Find the slope of the tangent to this curve at this point (1, 394)

HELP!!

2. how are you supposed to find the slope? ... straight power rule derivative, or derivative from the limit of a difference quotient?

3. The question is how to find the equation of the line tangent to a specified curve, $f(x)$, at a specified point, $(a, f(a))$. The answer is quick and easy:

$

y \ = \ f(a) \ + \ f'(a)(x-a),

$

where $f'(a)$ is the derivative of $f(x)$ evaluated at $a$.

4. so the derivative of 400-6x^2 is -12x

then where do I go?

5. Originally Posted by lava1980
so the derivative of 400-6x^2 is -12x

then where do I go?
Just follow the equation I posted, where $(a,f(a)) \ = \ (1, 394)$.

6. I'm thinking way too hard...

y = f(1) + f'(-12)...

but what's x?

Originally Posted by lava1980
so the derivative of 400-6x^2 is -12x

then where do I go?

7. Originally Posted by lava1980
I'm thinking way too hard...

y = f(1) + f'(-12)...

but what's x?
Here, x is a variable. It stays as x. Remember, you are trying to find the equation for a line.

8. Okay, that makes sense.

y =f(1) + f'(-12)(x-1)

What's after that?

Originally Posted by Varitron
Here, x is a variable. It stays as x. Remember, you are trying to find the equation for a line.

9. Originally Posted by lava1980
Okay, that makes sense.

y =f(1) + f'(-12)(x-1)

What's after that?
Actually evaluate f(1) and f'(-12), distribute and group like terms, and you are done.

10. But I have to find the slope of the tangent to the curve.

Does y = 394 + what?

Originally Posted by Varitron
Actually evaluate f(1) and f'(-12), distribute and group like terms, and you are done.

11. i think what he is trying to say is...

remember the derivative is going to be the slope...
if the slope is -12 then m=-12 and u have (x,y)-> (1, 394)

so solve for b in y=mx+b

394=-12(1)+b

then u have the equation for the line tangent to the point

correct me if im wrong, i just learned this last week =/

12. Originally Posted by lava1980
But I have to find the slope of the tangent to the curve.

Does y = 394 + what?
Firstly, I just realized one of your terms is wrong. It's not supposed to be f'(-12), it's f'(a) = f'(1) = -12. So, since f(1) = 394:

y = 394 - 12(x-1) = 394 - 12x + 12 = -12x + 406

The slope is obvious since it is in the form y = mx + b.

13. This seems correct, but I can't get other points on the graph to fit the equation...

f’(x) = -12x
if x = 1, then the slope = -12
if the slope is -12 then m=-12 and (x,y) > (1, 394)
so solve for b in y=mx+b
394=-12(1)+b
406 = b

y = -12x + 406

Any ideas?

Originally Posted by Khaali91
i think what he is trying to say is...

remember the derivative is going to be the slope...
if the slope is -12 then m=-12 and u have (x,y)-> (1, 394)

so solve for b in y=mx+b

394=-12(1)+b

then u have the equation for the line tangent to the point

correct me if im wrong, i just learned this last week =/

14. Originally Posted by lava1980
Here's the problem -

400-6x^2

Find the slope of the tangent to this curve at this point (1, 394)

HELP!!

only the point (1,394) will be on the graph of 400-6x^2 AND y=-12x+406

the tangent means it touches the graph 400-6x^2 once so no other point will be on it.