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# Use Graphs to Solve Quadratic Equations

## Identify x-intercepts of parabolas

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Practice Use Graphs to Solve Quadratic Equations
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Use Graphs to Solve Quadratic Equations

What if you had a quadratic function like \begin{align*}y = 2x^2 + 5x + 3\end{align*}? How could you graph it to find its roots? After completing this Concept, you'll be able to determine the number of solutions for a quadratic equation like this one and you'll find the solutions by graphing.

### Try This

Now that you’ve learned how to solve quadratic equations by graphing them, you can sharpen your skills even more by learning how to find an equation from the graph alone. Go to the page linked in the previous section, http://www.analyzemath.com/quadraticg/quadraticg.htm, and scroll down to section E. Read the example there to learn how to find the equation of a quadratic function by reading off a few key values from the graph; then click the “Click here to start” button to try a problem yourself. The “New graph” button will give you a new problem when you finish the first one.

### Guidance

Solving a quadratic equation means finding the \begin{align*}x-\end{align*}values that will make the quadratic function equal zero; in other words, it means finding the points where the graph of the function crosses the \begin{align*}x-\end{align*}axis. The solutions to a quadratic equation are also called the roots or zeros of the function, and in this section we’ll learn how to find them by graphing the function.

Identify the Number of Solutions of a Quadratic Equation

Three different situations can 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*}:

Looking at 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*}.

We can also find the solutions to the 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*}.

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 see that the graph touches the \begin{align*}x-\end{align*}axis at \begin{align*}x = 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 = 0\end{align*}, so \begin{align*}x = 1\end{align*}. Since the quadratic function is a perfect square, we get only one solution for the equation—it’s just the same solution repeated twice over.

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.

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.

So far we’ve found the solutions to quadratic equations using factoring. However, in real life very few functions factor easily. As you just saw, graphing a function gives a lot of information about the solutions. We can find exact or approximate solutions to a quadratic equation by graphing the function associated with it.

#### Example A

Find the solutions to the following quadratic equations by graphing.

a) \begin{align*}-x^2 + 3 = y\end{align*}

b) \begin{align*}-x^2 + x - 3 = y\end{align*}

c) \begin{align*}y = - x^2 + 4x - 4\end{align*}

Solution

Since we can’t factor any of these equations, we won’t be able to graph them using intercept form (if we could, we wouldn’t need to use the graphs to find the intercepts!) We’ll just have to make a table of arbitrary values to graph each one.

a)

\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.

b)

\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:

The graph curves up toward the \begin{align*}x-\end{align*}axis and then back down without ever reaching it. This means that the graph never intercepts the \begin{align*}x-\end{align*}axis, and so the corresponding equation has no real solutions.

c)

\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 graph just touches the \begin{align*}x-\end{align*}axis at \begin{align*}x = 2\end{align*}, so the function has a double root there. \begin{align*}x = 2\end{align*} is the only solution to the equation.

'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 or the vertex of the function.

#### Example B

Use a graphing calculator to analyze the graph of \begin{align*}y= x^2 - 20x + 35\end{align*}.

Solution

1. Graph the function.

Press the [Y=] button and enter “\begin{align*}x^2 - 20x + 35\end{align*}” next to \begin{align*}[Y_1=]\end{align*}. Press the [GRAPH] button. This is the plot you should see:

If this is not what you see, press the [WINDOW] button to change the window size. For the graph shown here, the \begin{align*}x-\end{align*}values should range from -10 to 30 and the \begin{align*}y-\end{align*}values from -80 to 50.

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 the cursor to the left of one of the roots and press [ENTER].

Move the cursor to the right of the same root and press [ENTER].

Move the cursor close to the root and press [ENTER].

The screen will show the value of the root. Repeat the procedure for the other root.

Whichever technique you use, you should get about \begin{align*}x = 1.9\end{align*} and \begin{align*}x = 18\end{align*} for the two roots.

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 value 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 the option ‘maximum’ if the vertex is a maximum or ‘minimum’ if the vertex is a minimum.

Move the cursor to the left of the vertex and press [ENTER].

Move the cursor to the right of the vertex and press [ENTER].

Move the 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.

Whichever method you use, you should find that the vertex is at (10, -65).

Solve Real-World Problems by Graphing Quadratic Functions

Here’s a real-world problem we can solve using the graphing methods we’ve learned.

#### Example C

Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the ball with respect to time is \begin{align*}y = -4.9t^2 + 48t\end{align*}, where \begin{align*}y\end{align*} is the height of the arrow in meters and \begin{align*}t\end{align*} is the time in seconds since Andrew shot the arrow. 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) = 117.6\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’s the graph of the function:

The roots of the function are approximately \begin{align*}x = 0\end{align*} sec and \begin{align*}x = 9.8\end{align*} sec. The first root tells us that the height of the arrow was 0 meters when Andrew first shot it. The second root says that it takes approximately 9.8 seconds for the arrow to return to the ground.

Watch this video for help with the Examples above.

### Vocabulary

• The solutions of a quadratic equation are often called the roots or zeros.

### Guided Practice

Find the solutions to \begin{align*} 2x^2 + 5x - 7 = 0\end{align*} by graphing.

Solution

Since we can’t factor this equation, we won’t be able to graph it using intercept form (if we could, we wouldn’t need to use the graphs to find the intercepts!) We’ll just have to make a table of arbitrary values to graph the equation.

\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*}
–3 \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)^2 + 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:

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 the equation.

### Practice

For 1-6, find the solutions of the following equations by graphing.

1. \begin{align*}x^2 + 3x + 6 = 0\end{align*}
2. \begin{align*}-2x^2 + x + 4 = 0\end{align*}
3. \begin{align*}x^2 - 9 = 0\end{align*}
4. \begin{align*}x^2 + 6x + 9 = 0\end{align*}
5. \begin{align*}10x - 3x^2 = 0\end{align*}
6. \begin{align*}\frac{1}{2}x^2 - 2x + 3 = 0\end{align*}

For 7-12, find the roots of the following quadratic functions by graphing.

1. \begin{align*}y = -3x^2 + 4x - 1\end{align*}
2. \begin{align*}y = 9 - 4x^2\end{align*}
3. \begin{align*}y = x^2 + 7x + 2\end{align*}
4. \begin{align*}y = -x^2 - 10x - 25\end{align*}
5. \begin{align*}y = 2x^2 - 3x\end{align*}
6. \begin{align*}y = x^2 - 2x + 5\end{align*}

For 13-18, use your graphing calculator to find the roots and the vertex of each polynomial.

1. \begin{align*}y = x^2 + 12x + 5\end{align*}
2. \begin{align*}y = x^2 + 3x + 6\end{align*}
3. \begin{align*}y = -x^2 - 3x + 9\end{align*}
4. \begin{align*}y = -x^2 + 4x -12\end{align*}
5. \begin{align*}y = 2x^2 - 4x + 8\end{align*}
6. \begin{align*}y = -5x^2 - 3x + 2\end{align*}
7. Graph the equations \begin{align*}y = 2x^2 - 4x + 8\end{align*} and \begin{align*}y = x^2 - 2x + 4\end{align*} on the same screen. Find their roots and vertices.
1. What is the same about the graphs? What is different?
2. How are the two equations related to each other? (Hint: factor them.)
3. What might be another equation with the same roots? Graph it and see.
8. Graph the equations \begin{align*}y = x^2 - 2x + 2\end{align*} and \begin{align*}y = x^2 - 2x + 4\end{align*} on the same screen. Find their roots and vertices.
1. What is the same about the graphs? What is different?
2. How are the two equations related to each other?
9. Phillip throws a ball and it takes a parabolic path. The equation of the height of the ball with respect to time is \begin{align*}y = - 16t^2 + 60t\end{align*}, where \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.
10. Use your graphing calculator to solve Ex. C. You should get the same answers as we did graphing by hand, but a lot quicker!

### Vocabulary Language: English

Double Root

Double Root

A solution that is repeated twice.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Roots

Roots

The roots of a function are the values of x that make y equal to zero.
Zero Product Rule

Zero Product Rule

The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero.
Zeroes

Zeroes

The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.
Zeros

Zeros

The zeros of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.