Solving Systems of Linear
and Quadratic Equations
Graphically
(also see Systems of Linear and Quadratic Equations)
A Linear Equation is an equation of a line.  
A Quadratic Equation is the equation of a parabola and has at least one variable squared (such as x^{2}) 

And together they form a System of a Linear and a Quadratic Equation 
A System of those two equations can be solved (find where they intersect), either:
 Using Algebra
 Or Graphically, as we will find out!
How to Solve Graphically
Easy! Plot both equations and see where they cross!
Plotting the Equations
We can plot them manually, or use a tool like the Function Grapher.
To plot them manually:
 make sure both equations are in "y=" form
 choose some xvalues that will hopefully be near where the two equations cross over
 calculate yvalues for those xvalues
 plot the points and see!
Choosing Where to Plot
But what values should we plot? Knowing the center will help!
Taking the quadratic formula and ignoring everything after the ± gets us a central xvalue:
Then choose some xvalues either side and calculate yvalues, like this:
Example: Solve these two equations graphically to 1 decimal place:
 y = x^{2} − 4x + 5
 y = x + 2
Find a Central X Value:
The quadratic equation is y = x^{2} − 4x + 5, so a = 1, b = −4 and c = 5
central x =  −b  =  −(−4)  =  4  = 2 
2a  2×1  2 
Now Calculate Values Around x=2
x 
Quadratic x^{2} − 4x + 5 
Linear x + 2 

0  5  2 
1  2  
2  1  
3  2  
4  5  
5  10  7 
(We only calculate first and last of the linear equation as that is all we need for the plot.)
Now Plot Them:
We can see they cross at about x = 0.7 and about x = 4.3
Let us do the calculations for those values:
x 
Quadratic x^{2} − 4x + 5 
Linear x + 2 

0.7  2.69  2.8 
4.3  6.29  6.2 
Yes they are close.
To 1 decimal place the two points are (0.7, 2.8) and (4.3, 6.2)
There Might Not Be 2 Solutions!
There are three possible cases:
 No real solution (happens when they never intersect)
 One real solution (when the straight line just touches the quadratic)
 Two real solutions (like the example above)
Time for another example:
Example: Solve these two equations graphically:
 4y − 8x = −40
 y − x^{2} = −9x + 21
How do we plot these? They are not in "y=" format!
First make both equations into "y=" format:
Linear equation is: 4y − 8x = −40
Quadratic equation is: y − x^{2} = −9x + 21
Now Find a Central X Value:
The quadratic equation is y = x^{2} − 9x + 21, so a = 1, b = −9 and c = 21
central x =  −b  =  −(−9)  =  9  = 4.5 
2a  2×1  2 
Now Calculate Values Around x=4.5
x 
Quadratic x^{2} − 9x + 21 
Linear 2x − 10 

3  3  4 
4  1  
4.5  0.75  
5  1  
6  3  
7  7  4 
Now Plot Them:
They never cross! There is no solution.
Real World Example
Kaboom!
The cannon ball flies through the air, following a parabola: y = 2 + 0.12x  0.002x^{2}
The land slopes upward: y = 0.15x
Where does the cannon ball land?
Let's fire up the Function Grapher!
Enter 2 + 0.12x  0.002x^2 for one function and 0.15x for the other.
Zoom out, then zoom in where they cross. You should get something like this:
By zooming in far enough we can find they cross at (25, 3.75)
Circle and Line
Example: Find the points of intersection to 1 decimal place of
 The circle x^{2} + y^{2} = 25
 And the straight line 3y  2x = 6
The Circle
The "Standard Form" for the equation of a circle is (xa)^{2} + (yb)^{2} = r^{2}
Where (a, b) is the center of the circle and r is the radius.
For x^{2} + y^{2} = 25 we can see that
 a=0 and b=0 so the center is at (0, 0),
 and for the radius r^{2} = 25 , so r = √25 = 5
We don't need to make the circle equation in "y=" form, as we have enough information to plot the circle now.
The Line
First put the line in "y=" format:
To plot the line, let's choose two points either side of the circle:
 at x = −6, y = (2/3)(−6) + 2 = −2
 at x = 6, y = (2/3)(6) + 2 = 6
Now plot them!
We can now see that they cross at about (4.8, 1.2) and (3.0, 4.0)
For an exact solution see Systems of Linear and Quadratic Equations