10.2: Quadratic Equations by Graphing
Learning Objectives
 Identify the number of solutions of quadratic equations.
 Solve quadratic equations by graphing.
 Find or approximate zeros of quadratic functions.
 Analyze quadratic functions using a graphing calculators.
 Solve realworld problems by graphing quadratic functions.
Introduction
In the last, section you learned how to graph quadratic equations. You saw that finding the \begin{align*}x \end{align*}intercepts of a parabola is important because it tells us where the graph crosses the \begin{align*}x\end{align*}axis. and it also lets us find the vertex of the parabola. When we are asked to find the solutions of the quadratic equation in the form \begin{align*}ax^2 + bx + c = 0\end{align*}, we are basically asked to find the \begin{align*}x \end{align*}intercepts of the quadratic function.
Finding the \begin{align*}x\end{align*}intercepts of a parabola is also called finding the roots or zeros of the function.
Identify the Number of Solutions of Quadratic Equations
The graph of a quadratic equation is very useful in helping us identify how many solutions and what types of solutions a function has. There are three different situations that occur when graphing a quadratic function.
Case 1 The parabola crosses the \begin{align*}x\end{align*}axis at two points.
An example of this is \begin{align*}y = x^2 + x  6\end{align*}.
We can find the solutions to equation \begin{align*}x^2 + x  6 = 0\end{align*} by setting \begin{align*}y = 0\end{align*}. We solve the equation by factoring \begin{align*}(x + 3) (x  2) = 0\end{align*} so \begin{align*}x = 3\end{align*} or \begin{align*}x = 2\end{align*}.
Another way to find the solutions is to graph the function and read the \begin{align*}x\end{align*}intercepts from the graph. We see that the parabola crosses the \begin{align*}x\end{align*}axis at \begin{align*}x = 3\end{align*} and \begin{align*}x = 2\end{align*}.
When the graph of a quadratic function crosses the \begin{align*}x\end{align*}axis at two points, we get two distinct solutions to the quadratic equation.
Case 2 The parabola touches the \begin{align*}x\end{align*}axis at one point.
An example of this is \begin{align*}y = x^2  2x + 1\end{align*}.
We can also solve this equation by factoring. If we set \begin{align*}y = 0\end{align*} and factor, we obtain \begin{align*}(x  1)^2\end{align*} so \begin{align*}x = 1\end{align*}.
Since the quadratic function is a perfect square, we obtain only one solution for the equation.
Here is what the graph of this function looks like. We see that the graph touches the \begin{align*}x\end{align*}axis at point \begin{align*}x = 1\end{align*}.
When the graph of a quadratic function touches the \begin{align*}x\end{align*}axis at one point, the quadratic equation has one solution and the solution is called a double root.
Case 3 The parabola does not cross or touch the \begin{align*}x\end{align*}axis.
An example of this is \begin{align*}y = x^2 + 4\end{align*}. If we set \begin{align*}y = 0\end{align*} we get \begin{align*}x^2 + 4 = 0\end{align*}. This quadratic polynomial does not factor and the equation \begin{align*}x^2 =4\end{align*} has no real solutions. When we look at the graph of this function, we see that the parabola does not cross or touch the \begin{align*}x\end{align*}axis.
When the graph of a quadratic function does not cross or touch the \begin{align*}x\end{align*}axis, the quadratic equation has no real solutions.
Solve Quadratic Equations by Graphing.
So far we have found the solutions to graphing equations using factoring. However, there are very few functions in real life that factor easily. As you just saw, graphing the function gives a lot of information about the solutions. We can find exact or approximate solutions to quadratic equations by graphing the function associated with it.
Example 1
Find the solutions to the following quadratic equations by graphing.
a) \begin{align*}x^2 + 3 = 0\end{align*}
b) \begin{align*}2x^2 + 5x  7 = 0\end{align*}
c) \begin{align*}x^2 + x  3 = 0\end{align*}
Solution
Let’s graph each equation. Unfortunately none of these functions can be rewritten in intercept form because we cannot factor the right hand side. This means that you cannot find the \begin{align*}x \end{align*}intercept and vertex before graphing since you have not learned methods other than factoring to do that.
a) To find the solution to \begin{align*}x^2 + 3 = 0\end{align*}, we need to find the \begin{align*}x \end{align*}intercepts of \begin{align*}y = x^2 + 3\end{align*}.
Let’s make a table of values so we can graph the function.
\begin{align*}x\end{align*}  \begin{align*}y = x^2 + 3\end{align*} 

\begin{align*}3\end{align*}  \begin{align*}y = (3)^2 + 3 = 6\end{align*} 
2  \begin{align*}y = (2)^2 + 3 = 1\end{align*} 
1  \begin{align*}y = (1)^2 + 3 = 2\end{align*} 
0  \begin{align*}y = (0)^2 + 3 = 3\end{align*} 
1  \begin{align*}y = (1)^2 + 3 = 2\end{align*} 
2  \begin{align*}y = (2)^2 + 3 = 1\end{align*} 
3  \begin{align*}y = (3)^2 + 3 = 6\end{align*} 
We plot the points and get the following graph:
From the graph we can read that the \begin{align*}x \end{align*}intercepts are approximately \begin{align*}x = 1.7\end{align*} and \begin{align*}x = 1.7\end{align*}.
These are the solutions to the equation \begin{align*} x^2 + 3 = 0\end{align*}.
b) To solve the equation \begin{align*}2x^2 + 5x  7 = 0\end{align*} we need to find the \begin{align*}x \end{align*}intercepts of \begin{align*}y = 2x^2 + 5 7.\end{align*}
Let’s make a table of values so we can graph the function.
\begin{align*}x\end{align*}  \begin{align*}y = 2x^2 + 5x  7\end{align*} 

\begin{align*}3\end{align*}  \begin{align*}y = 2(3)^2 + 5(3)  7 = 4\end{align*} 
2  \begin{align*}y = 2(2)^2 + 5(2)  7 = 9\end{align*} 
1  \begin{align*}y = 2(1)^2 + 5 (1)  7 = 10\end{align*} 
0  \begin{align*}y = 2(0)^0 + 5(0)  7 = 7\end{align*} 
1  \begin{align*}y = 2(1)^2 + 5(1)  7 = 0\end{align*} 
2  \begin{align*}y = 2(2)^2 + 5(2)  7 = 11\end{align*} 
3  \begin{align*}y = 2(3)^2 + 5(3)  7 = 26\end{align*} 
We plot the points and get the following graph:
Since we can only see one \begin{align*}x \end{align*}intercept on this graph, we need to pick more points smaller than \begin{align*}x = 3\end{align*} and redraw the graph.
\begin{align*}x\end{align*}  \begin{align*}y = 2x^2 + 5x  7\end{align*} 

\begin{align*}5\end{align*}  \begin{align*}y = 2(5)^2 + 5(5) 7 = 18\end{align*} 
4  \begin{align*}y =2(4)^2 + 5(4) 7 = 5\end{align*} 
Here is the graph again with both \begin{align*}x\end{align*}intercepts showing:
From the graph we can read that the \begin{align*}x\end{align*}intercepts are \begin{align*}x = 1\end{align*} and \begin{align*}x = 3.5\end{align*}.
These are the solutions to equation \begin{align*}2x^2 + 5x  7 = 0\end{align*}.
c) To solve the equation \begin{align*}x^2 + x  3 = 0\end{align*} we need to find the \begin{align*}x\end{align*}intercepts of \begin{align*}y = x^2 + x  3\end{align*}.
Let’s make a table of values so we can graph the function.
\begin{align*}x\end{align*}  \begin{align*}y = x^2 + x 3\end{align*} 

\begin{align*}3\end{align*}  \begin{align*}y = (3)^2 + (3) 3 = 15\end{align*} 
2  \begin{align*}y = (2)^2 + (2) 3 = 9\end{align*} 
1  \begin{align*}y = (1)^2 + (1) 3 = 5\end{align*} 
0  \begin{align*}y = (0)^2 + (0) 3 = 3\end{align*} 
1  \begin{align*}y = (1)^2 + (1) 3 = 3\end{align*} 
2  \begin{align*}y = (2)^2 + (2) 3 = 5\end{align*} 
3  \begin{align*}y = (3)^2 + (3) 3 = 9\end{align*} 
We plot the points and get the following graph:
This graph has no \begin{align*}x\end{align*}intercepts, so the equation \begin{align*}x^2 + x  3 = 0\end{align*} has no real solutions.
Find or Approximate Zeros of Quadratic Functions
From the graph of a quadratic function \begin{align*}y = ax^2 + bx + c\end{align*}, we can find the roots or zeros of the function. The zeros are also the \begin{align*}x\end{align*}intercepts of the graph, and they solve the equation \begin{align*}ax^2 + bx + c = 0\end{align*}. When the zeros of the function are integer values, it is easy to obtain exact values from reading the graph. When the zeros are not integers we must approximate their value.
Let’s do more examples of finding zeros of quadratic functions.
Example 2 Find the zeros of the following quadratic functions.
a) \begin{align*}y = x^2 + 4x  4\end{align*}
b) \begin{align*}y = 3x^2  5x\end{align*}
Solution
a) Graph the function \begin{align*}y = x^2 + 4x  4\end{align*} and read the values of the \begin{align*}x\end{align*}intercepts from the graph.
Let’s make a table of values.
\begin{align*}x\end{align*}  \begin{align*}y = x^2 + 4x  4\end{align*} 

\begin{align*}3\end{align*}  \begin{align*}y = (3)^2 + 4(3)  4 = 25\end{align*} 
2  \begin{align*}y = (2)^2 + 4(2) 4 = 16\end{align*} 
1  \begin{align*}y = (1)^2 + 4(1)  4 = 9\end{align*} 
0  \begin{align*}y = (0)^2 + 4(0)  4 = 4\end{align*} 
1  \begin{align*}y = (1)^2 + 4(1)  4 = 1\end{align*} 
2  \begin{align*}y = (2)^2 + 4(2)  4 = 0\end{align*} 
3  \begin{align*}y = (3)^2 + 4(3)  4 = 1\end{align*} 
4  \begin{align*}y = (4)^2 + 4(4)  4 = 4\end{align*} 
5  \begin{align*}y = (5)^2 + 4(5)  4 = 9\end{align*} 
Here is the graph of this function.
The function has a double root at \begin{align*}x = 2\end{align*}.
b) Graph the function \begin{align*}y = 3x^2  5x\end{align*} and read the \begin{align*}x\end{align*}intercepts from the graph.
Let’s make a table of values.
\begin{align*}x\end{align*}  \begin{align*}y = 3x^2  5x\end{align*} 

\begin{align*}3\end{align*}  \begin{align*}y = 3(3)^2 5(3) = 42\end{align*} 
2  \begin{align*}y = 3(2)^2  5 (2) = 22\end{align*} 
1  \begin{align*}y = 3(1)^2  5 (1) = 8\end{align*} 
0  \begin{align*}y = 3(0)^2  5 (0) = 0\end{align*} 
1  \begin{align*}y = 3(1)^2  5 (1) = 2\end{align*} 
2  \begin{align*}y = 3(2)^2  5 (2) = 2\end{align*} 
3  \begin{align*}y = 3(3)^2  5 (3) = 12\end{align*} 
Here is the graph of this function.
The function has two roots: \begin{align*}x = 0\end{align*} and \begin{align*}x \approx 1.7.\end{align*}
Analyze Quadratic Functions Using a Graphing Calculator
A graphing calculator is very useful for graphing quadratic functions. Once the function is graphed, we can use the calculator to find important information such as the roots of the function or the vertex of the function.
Example 3
Let’s use the graphing calculator to analyze the graph of \begin{align*}y = x^2  20 x + 35\end{align*}.
1. Graph the function.
Press the [Y=] button and enter “\begin{align*}X^220X+35\end{align*}” next to [\begin{align*}Y_1=\end{align*}].(Note, \begin{align*}X\end{align*} is one of the buttons on the calculator)
Press the [GRAPH] button. This is the plot you should see. If this is not what you see change the window size. For the graph to the right, we used window size of \begin{align*}\text{XMIN} = 10, \text{XMAX} = 30\end{align*} and \begin{align*}\text{YMIN} = 80, \text{YMAX} = 50\end{align*}. To change window size, press the [WINDOW] button.
2. Find the roots.
There are at least three ways to find the roots
Use [TRACE] to scroll over the \begin{align*}x\end{align*}intercepts. The approximate value of the roots will be shown on the screen.You can improve your estimate by zooming in.
OR
Use [TABLE] and scroll through the values until you find values of \begin{align*}Y\end{align*} equal to zero. You can change the accuracy of the solution by setting the step size with the [TBLSET] function.
OR
Use [2nd] [TRACE] (i.e. ‘calc’ button) and use option ‘zero’.
Move cursor to the left of one of the roots and press [ENTER].
Move cursor to the right of the same root and press [ENTER].
Move cursor close to the root and press [ENTER].
The screen will show the value of the root. For the left side root, we obtained \begin{align*}x = 1.9\end{align*}.
Repeat the procedure for the other root. For the right side root, we obtained \begin{align*}x = 18\end{align*}.
3. Find the vertex
There are three ways to find the vertex.
Use [TRACE] to scroll over the highest or lowest point on the graph. The approximate value of the roots will be shown on the screen.
OR
Use [TABLE] and scroll through the values until you find values the lowest or highest values of \begin{align*}Y\end{align*}.
You can change the accuracy of the solution by setting the step size with the [TBLSET] function.
OR
Use [2nd] [TRACE] and use option ‘maximum’ if the vertex is a maximum or option ‘minimum’ if the vertex is a minimum.
Move cursor to the left of the vertex and press [ENTER].
Move cursor to the right of the vertex and press [ENTER].
Move cursor close to the vertex and press [ENTER].
The screen will show the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values of the vertex.
For this example, we obtained \begin{align*}x = 10\end{align*} and \begin{align*}x = 65\end{align*}.
Solve RealWorld Problems by Graphing Quadratic Functions
We will now use the methods we learned so far to solve some examples of realworld problems using quadratic functions.
Example 4 Projectile motion
Andrew is an avid archer. He launches an arrow that takes a parabolic path. Here is the equation of the height of the ball with respect to time.
\begin{align*}y = 4.9t^2 + 48t\end{align*}
Here y is the height in meters and \begin{align*}t\end{align*} is the time in seconds. Find how long it takes the arrow to come back to the ground.
Solution
Let’s graph the equation by making a table of values.
\begin{align*}t\end{align*}  \begin{align*}y = 4.9t^2 + 48t\end{align*} 

0  \begin{align*}y = 4.9(0)^2 + 48(0) = 0\end{align*} 
1  \begin{align*}y = 4.9(1)^2 + 48(1) = 43.1\end{align*} 
2  \begin{align*}y = 4.9(2)^2 + 48(2) = 76.4\end{align*} 
3  \begin{align*}y = 4.9(3)^2 + 48(3) = 99.9\end{align*} 
4  \begin{align*}y = 4.9(4)^2 + 48(4) = 113.6\end{align*} 
5  \begin{align*}y = 4.9(5)^2 + 48(5) = 117.5\end{align*} 
6  \begin{align*}y = 4.9(6)^2 + 48(6) = 111.6\end{align*} 
7  \begin{align*}y = 4.9(7)^2 + 48(7) = 95.9\end{align*} 
8  \begin{align*}y = 4.9(8)^2 + 48(8) = 70.4\end{align*} 
9  \begin{align*}y = 4.9(9)^2 + 48(9) = 35.1\end{align*} 
10  \begin{align*}y = 4.9(10)^2 + 48(10) = 10\end{align*} 
Here is the graph of the function.
The roots of the function are approximately \begin{align*}x = 0 \ sec\end{align*} and \begin{align*}x = 9.8 \ sec\end{align*}. The first root says that at time 0 seconds the height of the arrow is 0 meters. The second root says that it takes approximately 9.8 seconds for the arrow to return back to the ground.
Review Questions
Find the solutions of the following equations by graphing.
 \begin{align*}x^2 + 3x + 6 = 0\end{align*}
 \begin{align*}2x^2 + x + 4 = 0\end{align*}
 \begin{align*}x^2  9 = 0\end{align*}
 \begin{align*}x^2 + 6x + 9 = 0\end{align*}
 \begin{align*}10x^2  3x^2 = 0\end{align*}
 \begin{align*}\frac{1} {2}x^2  2x + 3 = 0\end{align*}
Find the roots of the following quadratic functions by graphing.
 \begin{align*}y = 3x^2 + 4x  1\end{align*}
 \begin{align*}y = 9  4x^2\end{align*}
 \begin{align*}y = x^2 + 7x + 2\end{align*}
 \begin{align*}y = x^2  10x  25\end{align*}
 \begin{align*}y = 2x^2  3x\end{align*}

\begin{align*}y = x^2  2x + 5\end{align*} Using your graphing calculator
 Find the roots of the quadratic polynomials.
 Find the vertex of the quadratic polynomials.
 \begin{align*}y = x^2 + 12x + 5\end{align*}
 \begin{align*}y = x^2 + 3x + 6\end{align*}
 \begin{align*}y = x^2  3x + 9\end{align*}
 Peter throws a ball and it takes a parabolic path. Here is the equation of the height of the ball with respect to time: \begin{align*}y = 16t^2 + 60t\end{align*} Here \begin{align*}y\end{align*} is the height in feet and \begin{align*}t\end{align*} is the time in seconds. Find how long it takes the ball to come back to the ground.
 Use your graphing calculator to solve Ex. 5. You should get the same answers as we did graphing by hand but a lot quicker!
Review Answers
 No real solutions
 \begin{align*}x = 1.2, x = 1.87\end{align*}
 \begin{align*}x = 3, x = 3\end{align*}
 \begin{align*}x = 3\end{align*} double root
 \begin{align*}x = 0, x = 3.23\end{align*}
 No real solutions.
 \begin{align*}x = 0.3, x = 1\end{align*}
 \begin{align*}x = 1.5, x = 1.5\end{align*}
 \begin{align*}x = 6.7, x = 0.3\end{align*}
 \begin{align*}x = 5\end{align*} double root
 \begin{align*}x = 0, x = 1.5\end{align*}
 No real solutions.
 .
 .
 .
 \begin{align*}\text{time} = 3.75 \ second\end{align*}
 .
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