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Mathematics 19 Online
OpenStudy (anonymous):

This what I have so far...

OpenStudy (anonymous):

OpenStudy (anonymous):

f'(x)=4x x=2 f'(2)=4(2)=8 which is m x=2 f(2)=2(2)^2+3=11 which is y (8,11) y-11=8(x-2) y=8x-5 what do i do next to find y1 and y2?

OpenStudy (anonymous):

you need the equation for the line tangent to the graph. did you find that?

OpenStudy (anonymous):

huh?

OpenStudy (anonymous):

you found it right?

OpenStudy (anonymous):

the tangent line at (2,11) is \[y=8x-5 \]

OpenStudy (anonymous):

so if you want the y value at some other point, replace x by the x value given

OpenStudy (anonymous):

16-5=11

OpenStudy (anonymous):

yes, and you knew that one to begin with right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

since that is the point you used in the first place to find the line now replace x by -1 to find another point on the line

OpenStudy (anonymous):

-8-5=-13

OpenStudy (anonymous):

i guess. was the other x value -1?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

if so then you are right.

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