Question

Find the slope of y=-5-5x^2 at the given point (3,-50). Write the equation of the tangent line at the given point.

Answer 1

\(\bf y=-5-5x^2\qquad \cfrac{dy}{dx} = -10x\)

if you set x = 3, I'd think that'd be your slope

Answer 2

My professor gave an example for the class to work with. The example used the equation y=x^2 and gave point (2,4). In the end, its slope was, h+4.

Here is the example's process:

secant slope= dy/dx = ((2+h)^2-2^2)/h

So when I tried to use that process, my slope came out to be h+15.
Is that correct?

Answer 3

Oh I see, what I found was only the secant slope not the slope of the curve. But then what do I do with the secant slope to find the slope of the curve?

Answer 4

to get slope of curve from secant slope , we put the limit as h-> 0

Answer 5

how did u get h+15 ?

Answer 6

Sorry, I'm still confused.

Answer 7

Using the way the example did it, by doing ((3+h)^2-3^2)/h , I got h+15

Answer 8

slope of curve
\(\Large lim {h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\)

Answer 9

3 ?
isn't the Q have 5 ?

Answer 10

-5- 5(3+h)^2 - (-5-3^2)
will be the numerator, right ?

Answer 11

Oh! You're right, I was using the examples function y=x^2 rather than y=−5−5x^2.

Answer 12

so, try with -5-x^2 now ?

Answer 13

would the equation be (((-5-5x^2)+h)-(-5-5x^2))/h ?
Because I ended up with 1.

Answer 14

we replace 'x' by 'x+h'

so, the 1st term would be
-5- (x+h)^2

[and not ((-5-5x^2)+h) ]
got this ?

Answer 15

Ok, then wouldn't the first term be -5-5(x^2) ?

Answer 16

why would u say so ?

its -5 - 5(x+h)^2

and 2nd term = -5-5x^2

subtract those

Answer 17

f(3+h)-f(3) = -5- 5(3+h)^2 - (-5-3^2) = ... ?

Answer 18

These types of questions always get to me. I see now. Could you also just find (f(x+h)−f(x))/h (now I know the right way to do it) and then substitute in the 3 from that result?

Answer 19

And the 3 was from the x value, right?

Answer 20

yes, 3 is from the x-value of the point

f(3+h)-f(3) = -5- 5(3+h)^2 - (-5-3^2) = -5 - 5 (h^2+6h+9) +5+9

simplify this simple algebraic expression ?

Answer 21

= -5 - 5 (h^2+6h+9) +5+9

= -5h^2 -30h

= h (-5h-30)

this factored out h, cancels out with the h in the denominator , right ?

so you are left with
-5h-30

just put h=0 here to get the slope of the curve :)

Answer 22

and ask if any more doubts....

Answer 23

Thank you! :) I've looked all over the place trying to answer this question and it wasn't as hard as I thought.

Answer 24

correct! quite simple :)
welcome ^_^