How to graph multiple powered variables given an equation

[nivek516]

I was wanting to know how to graph \(\displaystyle x^3-x^2y=4xy-4y^2\). I was able to find out \(\displaystyle x^2(x-y)=4y(x-y)\) and \(\displaystyle x^2=4y\). Then, I became stuck. There has to be a way to graph this entire equation. Help please?

 

[stapel]

One method might be to solve for y in terms of x.

. . . . .4y[sup:ipb7d8t9]2[/sup:ipb7d8t9] + (-4x - x[sup:ipb7d8t9]2[/sup:ipb7d8t9])y + x[sup:ipb7d8t9]3[/sup:ipb7d8t9] = 0

Apply the Quadratic Formula to find y in terms of x. You'll get two functions (from the "plus / minus" on the square root). Graph each function.

Eliz.

 

[soroban]

Hello, nivek516!

I was wanting to know how to graph \(\displaystyle x^3-x^2y=4xy-4y^2\).
I was able to find out \(\displaystyle x^2(x-y)=4y(x-y)\) and \(\displaystyle x^2=4y\). Then I became stuck.
There has to be a way to graph this entire equation. Help please.

You were on your way . . .

\(\displaystyle \text{We have: }\:x^3-x^2y - 4xy + 4y^2 \:=\:0\\)

. . \(\displaystyle \text{which factors: }\:x^2(x-y) - 4y(x-y) \:=\:0 \quad\Rightarrow\quad (x-y)(x^2-4y) \:=\:0\)

\(\displaystyle \text{And we have }two\text{ equations: }\;\begin{array}{ccc} x-y&=&0 \\ x^2-4y&=& 0 \end{array}\)

\(\displaystyle \text{Any point that satisfies }either\text{ equation is on the graph.}\)

\(\displaystyle \text{So, the graph is comprised of two curves: }\;\begin{Bmatrix}\text{a line:} & y \:=\:x \\ \text{a parabola: } & y \:=\:\frac{1}{4}x^2 \end{Bmatrix}\)

And the graph looks like this:

   *            |            *
                |               *  
                |             *
    *           |           *  
                |         *
     *          |       *  *
      *         |     *   *
        *       |   *   *
           *    | *  *
. --------------*--------------
              * |
            *   |
          *     |
        *       |

 

[nivek516]

Thanks! I figured it out!