Question

Find the equation of the tangent to the graph at the indicated point (enter the equation in the y=mx+b form)
f(x)=x^2-2 ; x=6
y=?

Answer 1

You need a y right? Plug in the x=6 into the equation you have and see what it is.

Answer 2

Find the general slope by taking the derivative: \[\frac{dy}{dx} = (x^2 - 2)' = 2x\] The slope at x = 6 is 2(6) so 12 is your slope at that point. Then you can use the slope-intercept or point-slope form to solve for the equation of the line by using the slope 12x

Answer 3

plugging x=6 does not give you the y-intercept

Answer 4

How are you to find it then?

Answer 5

You plug in the 'x' and 'y' values for any point on the line, you already know a point which is the tangent point to the curve

Answer 6

soooo what would i do then? sorry a little confused

Answer 7

\[y = 12x + b \\ 34 = 12(6) + b \\ 34 = 72 + b \\ -38 = b\]

So y = 12x - 38 is your equation of the line tangent at the point

Answer 8

Alright. You need two things in order to create an equation. A slope and a point.

eye already found the slope to 12 so let's call slope m

m=12

now we need a point (x,y)

we have x=6 but we need a y. In order to find y and your point you have to plug in x=6 into the original equation\[f(x)=x^2-2\]

Answer 9

Can you tell me what you get after plugging in x=6?

Answer 10

how did you get y=34?

Answer 11

Wow, is OpenStudy lagging for anyone else

Answer 12

eyeh8maf was correct thank you all for helping

Answer 13

how did u get the y=34?

Answer 14

You are skipping steps and doing a lot of the work. Try to get the asker involved in the probem. You don't have to do all of the work.

Answer 15

@eyeh8maf You can answer that.

Answer 16

@eyeh8maf can you please?

Answer 17

please?;/

Answer 18

Sorry, I couldn't even load the page...this site is going very slow for me

Answer 19

I missed like the past 10 messages...to find the y-coordinate, you just plug in the x-coordinate into the original equation

Answer 20

\[f(x) = (6)^2 - 2 = 34\]

Answer 21

thank you very much!