Question

Find the derivative of 2x^2-13x+5 and use it to find an equation of the line tangent to the curve at x=3

Answer 1

Well, first you need to substitute x into f(x) to get a y value.
Then derive the f(x) to get the equation for the gradient)
(I'm trying to help you rather than do it for you lol)

Answer 2

okay I got -16 so its (3,-16) and then I take the limit of f(x)?
thank you by the way

Answer 3

Uh.. I'm not sure how you got -16. Substitute x=3 into 2x^2 - 13x + 5
So it will be (2 * (3^2)) - (13 * 3) + 5...

Answer 4

what are you getting? I still got -16

Answer 5

(2 * 9) - 29 + 5
18 - 29 + 5
-11 + 5
-6

Answer 6

13 x 3 is 39

Answer 7

OH. Jeez. That's embarrassing. *cough* let's just forget that happened, eh?
Alright, -16 is your y value. So the coordinates you want to find the equation to the tangent of are (3,-16).
Now derive your f(x) = 2x^2 - 13x + 5

Answer 8

lol its cool..
I got 2x-13 is that correct?

Answer 9

Almost, not quite.. When you derive a function...
f(x) = x^n
f'(x) = nx^(n-1)

You multiply the coefficient by the power and subtract one from the power. You missed something ;)

Answer 10

Oh yeah
okay so now we need to find what f'(3)?

Answer 11

Yeah. f'(x) is your gradient function, ie it will give you the gradient of the main function at any point. You want the gradient at (3,-16), so sub in 3 for x and that will give you the gradient at that point.
So what's your fixed derived equation? :)

Answer 12

f'(x)=4x-13
f'(3)=4(3)-13
f'(3)=-1
y+16=-1(x-3)
then I get....

Answer 13

Exactly! Now just expand and rearrange for y.

Answer 14

y=-1x-13

Answer 15

Yup! That's the equation for the tangent at that point.
You can double check with a graphics calculator or similar to make sure the lines meet at that point.

Answer 16

looks right :) Thank you so much

Answer 17

No worries dude :) have a great day

Answer 18

you too

Answer 19

Legend <3

Answer 20

Can I ask one more question?

Answer 21

Sure thing man, go ahead :)