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OpenStudy (anonymous):

Find the equation of the tangent line to the graph of p(x)=-x-3 at x=1.

OpenStudy (anonymous):

-1/2x^2 - 3x

OpenStudy (anonymous):

How?

OpenStudy (anonymous):

the only way to get -x as a derivative is -1/2x^2 and the only way to get -3 as a derivative is -3x

OpenStudy (anonymous):

i dont know the specific rule/law/theorem though

OpenStudy (anonymous):

I still don't understand.

OpenStudy (anonymous):

do you know how to get a derivative going forward

OpenStudy (anonymous):

is this really \[p(x)=-x-3\]?

OpenStudy (anonymous):

f'(x) = -x - 3

OpenStudy (anonymous):

oooooooooooooh

OpenStudy (anonymous):

that is not what it says. it says Find the equation of the tangent line to the graph of p(x)=-x-3 at x=1.

OpenStudy (anonymous):

my bad, sorry, im not that far yet in math then lol

OpenStudy (anonymous):

and since that is a line itself, the tangent line is the same as that line

OpenStudy (anonymous):

So the answer would be -x-3?

OpenStudy (anonymous):

i believe the answer is -1 because the derivative of p(x)= -x-3 is -1

OpenStudy (anonymous):

-1 is the slope for the line you need a point so plug in x=1 in the original equation and then do point slope

OpenStudy (anonymous):

if the question is really the one you asked, then you are asked to find the equation of a line tangent to a line

OpenStudy (anonymous):

sorry i was so confused before

OpenStudy (anonymous):

a rather silly question, but it is the line itself. that is all

OpenStudy (anonymous):

so what about this then...Find the equation of the tangent line \to the graph of \[h(x)=2x^3-4x^2 at x=-2\]

OpenStudy (anonymous):

now that is a more reasonable question. 1) take the derivative 2) evaluate the derivative at x = -2 to get the slope 3) evaluate the function at x - -2 4) use the point - slope formula

OpenStudy (anonymous):

first find the derivative of the equation - 6x^2 - 8x, that is the slope at any given point, plug in -2 for the x value and that is the slope of the line then, plug in -2 into the original equation, that is the y value of the point (-2,y), use that as your point for point slope

OpenStudy (anonymous):

So help me work this out... 1)

OpenStudy (anonymous):

and it is 6x^2 -8x, not -6 sorry

OpenStudy (anonymous):

6x^2-8x?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

then, plug in the -2

OpenStudy (anonymous):

okay then 2) 6(-2)^2-8(-2)

OpenStudy (anonymous):

you plug in -2 into the orignial equation to get the y-coordinate which is -32. then you find the derivative using the power rule (6x^2-8x) and then plug in x=-2 into the derived equation to get the slope at that particular point. the equation would then be y+32=40(x+2)

OpenStudy (anonymous):

24+16

OpenStudy (anonymous):

yeah, 40

OpenStudy (anonymous):

that is the slope, now plug -2 into the original equation

OpenStudy (anonymous):

k

OpenStudy (anonymous):

then what?

OpenStudy (anonymous):

so now i have -32 and 40

OpenStudy (anonymous):

your point to plug in to point slope.... \[y-y _{1} = m(x-x _{1})\] so your point is (-2, -32) and your slope or m is 40

OpenStudy (anonymous):

So plug it in for me?

OpenStudy (anonymous):

Please

OpenStudy (anonymous):

i dont have a calculator, and most teachers let you leave it as point slope so it is \[y+32=40(x+2)\]

OpenStudy (anonymous):

perfect...thank you

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