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Find a set of parametric equations $x(m)$, $y(m)$ that represent the graph of $y= 2x^2-3x+1$ such that the parameter $m$ is the slope $m = \frac{dy}{dx}$ at the point $(x,y)$.

I am guessing I need first convert the polynomial into parametric first? Honestly, I have no clue how to even start this. I've looked for hours on the internet but can't find an explanation that I understand.

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  • $\begingroup$ (1) Find $dy/dx$ in terms of $x$. Call the result $m$. (2) Solve for $x$ in terms of $m$. (3) Replace $x$ by the result of step (2) in the original equation to get $y$ in terms of $m$. Done. $\endgroup$ – quasi Feb 6 '17 at 18:23
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Since $m=y'=4x-3$ then $$x=\frac{m+3}{4}$$ Then the graph is $$(x,y)=(x,2x^2-3x+1)=\left(\frac{m+3}4,\frac{m^2-6m+9}8-\frac{3m+9}4+1\right)$$ or $$(x(m),y(m))=\left(\frac{m+3}4,\frac{m^2-12m-1}8\right)$$

Note that it is crucial that you can solve for $x$ in the equation of the derivative, so the polynomial should have a degree at most $2$. Or, at least, the derivative of the polynomial should be an injective function.

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