Question

given f(x)=(x^2)-2x
a: Find f ' (x)
b: Find the slopes of the lines tangent to the graph of f at x=-2, -1, and 2
c: Graph f, and sketch in the tangent lines at x=-2, -1, and 2

Answer 1

a) Use power rule.

b) It asks you to find f'(-2), f'(-1), and f'(2).

Answer 2

I'm taking and online course and am having trouble understanding how to do those things, thats my problem =/

Answer 3

2x-2? for A?

Answer 4

a)

like RD⁴² is saying you use the power rule

so if f(x) = x^3 for example, then f ' (x) = 3x^2

in general, if f(x) = x^n, then f ' (x) = n*x^(n - 1)

Answer 5

so good, you would derive f(x) = x^2 - 2x to get f ' (x) = 2x - 2

Answer 6

just saw your response lol

Answer 7

now you would use this for part b

Answer 8

f ' (x) = 2x - 2

f ' (-2) = 2(-2) - 2

f ' (-2) = -4 - 2

f ' (-2) = -6

This means that the slope of the tangent line at x = -2 (on f(x)) is -6

Answer 9

you would repeat for x = -1 and x = 2

Answer 10

Okay! Formulas are one thing but putting them on a graph?! Theres a real issue haha! You are an amazing person thanks so much for taking the time! I wish you were a tutor!

Answer 11

lol I am a tutor (sorta)

to graph f(x)=(x^2)-2x, you can plot a bunch of points or you can use a graphing calculator

Answer 12

Well an online tutor because I'd seriously pay you, I actually understand things when you go through them with me!

Answer 13

here is the graph of f(x)=x^2-2x shown in red

Answer 14

now locate x = -2 and plot the point (-2, 8) on the graph

Answer 15

you can see that when x = -2, y is...

y = x^2 - 2x

y = (-2)^2 - 2(-2)

y = 4 + 4

y = 8

So that algebraically explains why the point (-2, 8) lies on the graph

Answer 16

from there, you draw a tangent line that goes through (-2, 8)

you do this by drawing a line with a slope of -6 that goes through (-2, 8)

so start at (-2, 8) and go down 6 and to the right 1 to get to the next point (-1, 2)

the added green line is the tangent line at x = -2

Answer 17

does that make sense?

Answer 18

It really does...like I said, you're awesome! And you dont make me feel like an idiot!

Answer 19

and I do offer paid tutoring, but tutoring I do on here is free since this site is free

Answer 20

sweet, glad it's making more sense

and no definitely not, you're learning calculus here lol

Answer 21

Wheat site do you offer paid tutoring and how much?

Answer 22

I just offer it through email or chat

no official site really (at the moment I guess)

Answer 23

If you have any availability I am really interested!

Answer 24

let me know if you still need help on this problem, but for the most part it sounds like you got this one (which is great)