I feel like this is an easy question but I'm having trouble Please Help!
For each function, Find the points on the graph at which the tangent Line has slope 1
y=1/3 x^3+2x^2+2x
Thank you in advance!
My Differentiation techniques are not working.
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calculus
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0I cant even get the y' of the problem I have it set up as 1= x^2+4x+2 and then I -1 so its x^2+4x+1 but that doesn't factor so I'm stuck – 2017-01-27
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0We can add $2$ to each side of this equation to get $3 = x^2 + 4x + 4$. See if this helps! – 2017-01-27
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2Try the quadratic formula. – 2017-01-27
1 Answers
1
after the power rule we have $$y'=x^2+4x+2$$ and now solve the equation $$x^2+4x+2=1$$ for $x$ where the power rule is given as $$(x^n)'=nx^{n-1}$$
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4Hi, we're trying to elicit some shown work from the OP so we know where to help. I'm not sure posting the solution is equally as beneficial for them. – 2017-01-27