Graphing Quadratic Systems

Solving Quadratic Systems by Graphing

You may remember solving linear systems of equations by graphing from Algebra 1. Solving quadratic systems follows a very similar process! It would be a good idea to brush up on your graphing skills before diving in as well.

In short, a solution to a system of equations is a point, $(x,y)$, where all of the functions in the system cross. Algebraically solving a system normally involves solving one of your equations to find value(s) for $x$, then plugging that value into the other equation to find the value(s) for $y$. Graphing an entire system gives you the solutions by visually showing you the crossing points of the various curves.

The fourth graph shows two functions that do not cross. This system of equations would not have any real solutions. Let's look at each possible case of solutions for different equations.

No Solutions

This may look familiar as it is the same system as the answer to the problem given above! This graph shows the following system of equations: $$\begin{cases} y&=x^2\\y&=-x-1 \end{cases}$$ The two functions do not cross so there is no solution.

One Solution

This depicts the following system of equations: $$\begin{cases} y&=x^2\\ y&=0 \end{cases}$$ They only touch at one point, $(0,0)$. The values of $x=0$ and $y=0$ make the only solution to this system. We can check this by plugging in $0$ for $x$ and $y$ and seeing if the equations in the system are true. $$\begin{align}\\y&=x^2\\ 0&=0^2\\ 0&=0\\ \\ y&=0\\ 0&=0\end{align}$$ The point $(0,0)$ makes each equation a true statement.

Two Solutions

This graph shows the following system: $$\begin{cases}y&=x^2+2x+1\\y&=3x+3\end{cases}$$ These two functions have two crossing points: $(-1,0)$ and $(2,9)$. Let's check to see if these points each equation true $$\begin{align}y&=x^2+2x+1\\ 9&=2^2+2(2)+1\\ 9&=9\\ y&=3x+3\\ 9&=3(2)+3\\ 9&=9\\ \\ y&=x^2+2x+1\\ 0&=(-1)^2+2(-1)+1\\ 0&=1-2+1\\ 0&=0\\ \\ y&=3x+3\\ 0&=3(-1)+3\\ 0&=0\end{align}$$ All of the points make the equations true, confirming that this system has $2$ solutions.

Infinite Solutions

There can be infinite solutions if the two parabolas are equal to each other. Let's look at the following system: $$\begin{cases}y&=x^2+2x+1\\y&=(x+1)^2\end{cases}$$

These two graphs are overlapping at all points on the curve, so it has an infinite number of solutions.

In any of these cases there are $3$ steps to solving systems by graphing:

1. Graph the functions
2. Find the crossing points
3. Check to see if the crossing points are valid

(Video) Solving Nonlinear Systems of Equations

by HCCMathHelp

Here's a good video by HCCMathHelp showing you how to solve a system of nonlinear equations. One of the methods is graphing which hes shows you how to do.

He starts with the following system of equations: $$\begin{cases}y=x^2+x-1\\y=-x+2\end{cases}$$ Both of these equations start with $y=$, so they must also be equal to each other. We can set them equal to each other and solve for $x$. $$\begin{align}x^2+x-1&=-x+2\\ x^2+2x-3&=0\\ (x+3)(x-1)&=0\\ x&=-3, 1 \end{align}$$ Notice that we have two solutions. This is because one of our equations is a quadratic. We can confirm this by graphing the two functions and counting the number of crossing points.

We can see two very clear crossing points on the graph.

We now need to plug the newly found $x$ values into one of our functions to find our $y$ values. Let's start with $x=1$. $$\begin{align}y&=x^2+x-1\\ y&=(1)^2+1-1\\ y&=1\\ \\ y&=-x+2\\ y&=-(1)+2\\ y&=1\end{align}$$ The value $x=1$ gives us a solution of the point $(1,1)$. Now we can check with $x=-3$. $$\begin{align}y&=x^2+x-1\\ y&=(-3)^2+(-3)-1\\ y&=9-4\\ y&=5\\ \\ y&=-x+2\\ y&=-(-3)+2\\ y&=5\end{align}$$

He then explains that by accurately graphing the functions (either by hand or using a calculator) you can find the solutions by finding the exact points where the functions cross!

(Video) Solving a Nonlinear System of Equations

by Angiewvc

Here's a video by Angiewvc showing you how to solve a system of nonlinear equations.

She starts by sketching a graph for the following system: $$\begin{cases}y=x^2-8\\y=2x-5\end{cases}$$

This system will have $2$ solutions as it crosses the graph twice.

She points out that both equations are definitions of $y$, so we can set them equal to each other in order to solve for $x$. $$\begin{align}x^2-8&=2x-5\\ x^2-2x-3&=0\\ (x-3)(x+1)&=0\\ x&=3,-1 \end{align}$$

You can check these $x$ values with your graph. You can see crossing points on the graph at $x=-1$ and $x=3$.

The last step is to plug your newly found $x$ values into one of your equations to find the $y$ values. She uses $y=x^2-8$ to do this for the first $x$, and $y=2x-5$ for the second $x$.

You can also check to see if this is correct by plugging points into the opposite equations. Let's check this. $$\begin{align}y&=x^2-8\\ y&=(-1)^2-8\\ y&=1-8\\ y&=-7\end{align}$$ We get the same $y$ value corresponding with $x=-1$, so we have a correct set of points!.