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anonymous
 4 years ago
This what I have so far...
anonymous
 4 years ago
This what I have so far...

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0f'(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) y11=8(x2) y=8x5 what do i do next to find y1 and y2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you need the equation for the line tangent to the graph. did you find that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the tangent line at (2,11) is \[y=8x5 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so if you want the y value at some other point, replace x by the x value given

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, and you knew that one to begin with right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess. was the other x value 1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if so then you are right.
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