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This what I have so far...

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

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Other answers:

huh?
you found it right?
the tangent line at (2,11) is \[y=8x-5 \]
so if you want the y value at some other point, replace x by the x value given
16-5=11
yes, and you knew that one to begin with right?
yes
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
-8-5=-13
i guess. was the other x value -1?
yes
if so then you are right.

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