Question

Find the equation of the tangent line to the curve f(x) = 2x2 – x at x = 1.

A. T(x) = 2 + 3(x – 1)
B. T(x) = 4 + 3(x – 1)
C. T(x) = 1 + 3(x – 1)
D. T(x) = 1 – 3(x – 1)

Answer 1

@zepdrix can u help?

Answer 2

\[\large f(x)=2x^2-x\]Were you able to find the derivative of \(f(x)\) ok?
We'll need that for the `slope` of our tangent line.

Answer 3

no i haven't yet

Answer 4

Why not? D: Don't you remember the power rule? c:

Answer 5

i do

Answer 6

\[\large f(x)=2x^2-x^1\]Applying the power rule to each term,
~Bringing the power down as a coefficient and,
~decreasing the power by 1,\[\large f'(x)=2\cdot2x^{1}-1\cdot x^{0}\]\[\large f'(x)=4x-1\]

Answer 7

the i substitute the one right?

Answer 8

The equation of our tangent line, in slope-intercept form, will look like this.\[\large y=\color{royalblue}{m}x+b\]Where \(m\) is the slope of the tangent line at x=1.
In other words,\[\large \color{royalblue}{m}=f'(1)\]

Yes good, plug in x=1, and solve for a value, that will be our m!

Answer 9

okay so the m will be 3?

Answer 10

yup sounds good.

Answer 11

thanks

Answer 12

Closed?? XD You know how to solve it from here? :D

Answer 13

i have an idea..lol

Answer 14

but not 100% on it

Answer 15

Hmm it looks like your multiple choice answers were generated by using `Point-slope Form`.\[\large y-y_o=m(x-x_o)\]

So maybe we should use that :O Either way, no big deal.
Plugging x=1 into your original \(f(x)\) will give you a y value, what'd u get for that? :D

Answer 16

i got 3

Answer 17

\[\large f(1)=2(1)^2-(1)\]Hmmmm I think it's 1.

Answer 18

okay

Answer 19

So this is our \(\large (x_o,y_o)\), the point \(\large(1,1)\).
We'll plug that into the formula and it should look very similar to one of our options.

\[\large y-y_o=m(x-x_o) \qquad \rightarrow \qquad y-1=3(x-1)\]

Answer 20

so it will be D?

Answer 21

Hmmm I think it's probably C.

Answer 22

okay