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Find the slope of the line tangent to the curve at x. y = -7x2 + 5x; x = 11
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When you see slope think derivative? The derivation is the slope at every point of the curve. y'=? Then evaluate y' at x=11
derivative*
\[y = -7x ^{2} + 5x\] \[x = 11\] Take the derivative, so \[y' =-14x + 5 \] Then plug in 11, and the slope at x = 11 is -14(11) + 5 = -149
thank you!! i saw an example like that on the internet but i didnt think it would be that easy, thank you so much
Very welcome!
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i just posted another question, if u could at least explain to me how it works...
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