Let f(x) = 3 x^2 - 8. Find an equation of the line tangent to the graph of f at the point where x = -3.

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Let f(x) = 3 x^2 - 8. Find an equation of the line tangent to the graph of f at the point where x = -3.

Mathematics
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what is the equation first?
first find the second coordinate by replacing x by -3
the equation would be 3 -3^2-8

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y=14
i get \[y=3\times 9-8=27-8=19\] so the point on your line is \[(-3,19)\] then find the slope. the derivative is \[y'=6x\] and replacing x by -3 gives you \[-18\] so you have a slope and a point, use point slope formula
y=mx+b
\[y-19=-18(x+3)\] etc
Well imagine that. The same answer as when I helped you with the same problem.
yeah but its not coming out right
What do you mean it's not coming out right?
I plugged in y=-18x+b and y-19=-18(x+3) and on the website Im supposed to submit is saying that it is incorrect
Well plug in y = -18x-35
You told me before that your book said the answer was y = -18x+b
I know but the website wasnt taking that answer
Did you put in y = -18x-35 yet?
but the website just took that answer so its right and the book didnt want me to simplify
thanx
Well i think I told you that was the answer 1 1/2 hours ago.
yw
what a waste of 1 1/2 hours

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