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### Section 1-5 : Functions of Several Variables

9. Identify and sketch the traces for the following function.

$4z + 2{y^2} - x = 0$

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We have two traces. One we get by plugging $$x = a$$ into the equation and the other we get by plugging $$y = b$$ into the equation. Here is what we get for each of these.

\begin{align*}x = a: & \hspace{0.25in}4z + 2{y^2} - a = 0 & \to \hspace{0.25in} & z = - \frac{1}{2}{y^2} + \frac{a}{4}\\ y = b: & \hspace{0.25in}4z + 2{b^2} - x = 0 & \to \hspace{0.25in} & x = 4z + 2{b^2}\end{align*} Show Step 2

Okay, we’re now into a realm that many students have issues with initially. We no longer have equations in terms of $$x$$ and $$y$$. Instead we have one equation in terms of $$x$$ and $$z$$ and another in terms of $$y$$ and $$z$$.

Do not get excited about this! They work the same way that equations in terms of $$x$$ and $$y$$ work! The only difference is that we need to make a decision on which variable will be the horizontal axis variable and which variable will be the vertical axis variable.

Just because we have an $$x$$ doesn’t mean that it must be the horizontal axis and just because we have a $$y$$ doesn’t mean that it must be the vertical axis! We set up the axis variables in a way that will be convenient for us.

In this case since both equation have a $$z$$ in them we’ll let $$z$$ be the horizontal axis variable for both of the equations.

So, given that convention for the axis variables this means that for the $$x = a$$ trace we’ll have a parabola that opens to the left with vertex at $$\left( {\frac{a}{4},0} \right)$$ and for the $$y = b$$ trace we’ll have a line with slope of 4 and an $$x$$-intercept at $$\left( {0,2{b^2}} \right)$$.

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Below is a sketch for each of the traces.  