3. Solving Quadratic Equations by Graphing
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Chapter 8
3.
Solving Quadratic Equations by Graphing
| Student Learning Objectives: |
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Why
Solving quadratic equations graphically offers a visual method to understand the solutions. By plotting the related function of the equation on a graph, one can identify the x-intercepts, which represent the solutions. This method is particularly useful in real-world scenarios, such as determining the height of a rocket at a specific time or understanding the flight path of an arrow. The graphical representation provides clarity on whether the equation has one, two, or no solutions based on the points of intersection with the x-axis. Additionally, the axis of symmetry and vertex offer insights into the characteristics of the quadratic function. This approach not only aids in solving equations but also enhances comprehension of quadratic functions in various contexts.
Lesson Settings & Tools
| | 9 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Quadratic Equations by Graphing
Slide of 9
The graph of an equation in two variables shows the set of all possible solutions. In this lesson, equations involving quadratic expressions will be solved by graphing.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Finding Solutions of Quadratic Equations
Consider a quadratic function in standard form. f(x) = 3/4x^2 -3x Recall the steps followed to graph a quadratic function in standard form. Use the applet to draw the graph.
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Discussion
Quadratic Equation
A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is 2. A quadratic equation can be written in standard form as follows.
ax^2+bx+c=0
Here, the leading coefficient a is non-zero, which guarantees that the x^2-term is present. Solving a quadratic equation means finding the zeros of the related quadratic function. Therefore, quadratic equations can have at most two solutions. c|c Quadratic Equation & Related Function [0.4em] ax^2+bx+c=0 & y = ax^2+bx+c
This type of equation can be solved using several methods, such as graphing and factoring.Method
Solving a Quadratic Equation Graphically
A quadratic equation can be solved graphically by drawing the parabola that corresponds to the related quadratic function and identifying its zeros. Consider the following equation as an example. 2x^2+3x-5=5x+7 These four steps can be followed to find the solutions of the equation.
1
Write the Equation in Standard Form
expand_more If neither side of the equation equals 0, rearrange it so that all terms are on the same side.
2x^2+3x-5=5x+7 ⇕ 2x^2+3x-5-5x-7=0
If there is more than one term with the same degree, simplify the equation by combining like terms.
2x^2+3x-5-5x-7=0 ⇕ 2x^2-2x-12=0
The equation is in standard form now.
2
Graph the Related Function
expand_more The side with the variables can now be seen as a function f(x). Quadratic Equation 2x^2-2x-12=0 [0.6em]
Related Function f(x) = 2x^2-2x-12
To graph the quadratic function in standard form, its characteristics should be identified first.
| f(x) = 2x^2-2x-12 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | y-intercept |
| a > 0 ⇒ upward | (1/2,- 25/2 ) | x= 1/2 | (0,- 12) |
In addition to these points, the reflection of the y-intercept across the axis of symmetry is at (1,- 12), which is also on the parabola. Now, the graph can be drawn.
3
Identify Any Points With y-coordinate 0
expand_more Now, find all the points on the graph that have a y-coordinate equal to 0.
4
Identify the x-coordinates of Those Points
expand_more The x-coordinates of the identified points solve the equation f(x)=0.
In the example, the x-coordinates of the points where the curve intercepts the x-axis are - 2 and 3. Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into f(x) and evaluate the expression. Verifying the solution x = - 2 is done by evaluating f(- 2).
f(x) = 2x^2-2x-12
Substitute
x= - 2
f( - 2) = 2 ( - 2)^2-2( - 2)-12
▼
Evaluate right-hand side
CalcPow
Calculate power
f(- 2) = 2(4)-2(- 2)-12
Multiply
Multiply
f(- 2) = 8-2(- 2)-12
MultPosNeg
a(- b)=- a * b
f(- 2) = 8-(- 4)-12
SubNeg
a-(- b)=a+b
f(- 2) = 8+4-12
AddSubTerms
Add and subtract terms
f(- 2) = 0
When x=- 2, the value of f(x) is 0. Therefore, x=- 2 is an exact solution. The solution x = 3 can now be verified by following the same procedure.
| Solution | Substitute | Evaluate |
|---|---|---|
| x= - 2 | f( - 2) = 2 ( - 2)^2-2( - 2)-12 | f(- 2)=0 ✓ |
| x= 3 | f( 3) = 2 ( 3)^2-2( 3)-12 | f(3)=0 ✓ |
Since f(3)=0, x=3 is also a solution. Therefore, the equation 2x^2+3x-5=5x+7 has two solutions. x=- 2 and x=3
Concept
Number of Solutions of a Quadratic Equation
The solutions of a quadratic equation can be interpreted graphically as the zeros of the related quadratic function. Therefore, the number of solutions of a quadratic equation is the same as the number of zeros of the related function. c|c Quadratic Equation & Quadratic Function [0.5em] ax^2+bx+c=0 & y=ax^2+bx+c If the function has two zeros, the equation ax^2+bx+c=0 has two solutions. Similarly, if the function has one zero, the equation has one solution. Finally, if the function does not have any zeros, the equation has no real solutions.
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Pop Quiz
Using a Graph to Find the Number of Solutions of a Quadratic Equation
For each quadratic equation, draw its related quadratic function to determine the number of solutions.
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Example
Finding the Time When a Rocket Reaches a Certain Height
Paulina models the flight of a rocket she built using a quadratic function, where h is the height of the rocket in meters after t seconds.
a After how many seconds does the rocket reach a height of 192 meters?
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b After how many seconds does the rocket reach a height of 196 meters?
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Hint
a Set the quadratic function equal to 192. Then, rewrite it in standard form to solve by graphing.
b Set the quadratic function equal to 196. Then, rewrite it in standard form to solve by graphing.
Solution
a To find after how many seconds the rocket is at a height of 192 meters, 192 will be substituted for h into the given quadratic function.
Now the function related to the equation can be written.
Equation - 16t^2 + 112t -192 = 0 ⇓ Related Function f(t)=- 16t^2 + 112t -192
The solutions to the equation are the x-intercepts of the function's graph. To graph the related quadratic function, its characteristics must be first determined.
The parabola intersects the x-axis twice. The points of intersection are ( 3,0) and ( 4,0). 
h = - 16t^2 + 112t ⇓ 192 = - 16t^2 + 112t This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form.
192 = - 16t^2 + 112t
SubEqn
LHS-192=RHS-192
0 = - 16t^2 + 112t -192
RearrangeEqn
Rearrange equation
- 16t^2 + 112t -192 = 0
| f(t)=- 16t^2 + 112t -192 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | y-intercept |
| a < 0 ⇓ downward | (3.5,4) | x= 3.5 | (0,- 192) |
In addition to the vertex and the y-intercept, the reflection of the y-intercept across the axis of symmetry, which is at (7,- 192), is also on the parabola. With this information, the graph of the function can be drawn.
Therefore, the equation - 16t^2 +112t-192=0 has two solutions, t_1= 3 and t_2= 4. - 16t^2 +112t-192=0 ↙ ↘ t_1 = 3 t_2 = 4
These solutions mean that the rocket is at a height of 192 meters after 3 and after 4 seconds. b Following the same steps as above, the time at which the rocket is at 196 meters can be calculated.
h = - 16t^2 + 112t ⇓ 196 = - 16t^2 + 112t This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form.
196 = - 16t^2 + 112t
SubEqn
LHS-196=RHS-196
0 = - 16t^2 + 112t -196
RearrangeEqn
Rearrange equation
- 16t^2 + 112t -196 = 0
Now the function related to the equation can be written. Equation - 16t^2 + 112t -196 = 0 ⇓ Related Function g(t)=- 16t^2 + 112t -196 Again, the solutions to the equation are the x-intercepts of the parabola. Identify the characteristics of the function to be able to draw it.
| g(t)=- 16t^2 + 112t -196 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | y-intercept |
| a < 0 ⇓ downward | (3.5,0) | x= 3.5 | (0,- 196) |
The reflection of the y-intercept across the axis of symmetry is at (7,- 196), which is also on the parabola. Now, the graph can be drawn.
This time, there is only one x-intercept.
The point of intersection is ( 3.5,0). Therefore, the equation - 16t^2 +112t-196=0 has one solution, t= 3.5. - 16t^2 +112t-196=0 ↓ t = 3.5 This solution means that the rocket is at a height of 196 meters after 3.5 seconds.
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Example
Modeling Areas of Geometric Figures With Quadratic Equations
Tadeo wants to form a square and a right triangle whose areas are equal. The side lengths of the geometric shapes are expressed in the diagram.
a Write a quadratic equation in standard form to model the situation and solve it by graphing.
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b Let f(x) and g(x) be the functions representing the areas of the square and the triangle, respectively. Express these functions in terms of x.
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Hint
a The area of a square equals its side length squared. The area of a right triangle is half the product of its legs.
b The function for the area of the triangle is a linear function. To draw its graph, make a table of values.
Solution
a Using the given diagram, an expression for the area of each shape can be written.
The area of the square is the square of 0.5x, and the area of the triangle is half the product of 2 and 0.5x+2.
| Area of Square | Area of Triangle |
|---|---|
| (0.5x)^2 | 1/2 ( 2 ) (0.5x+2) |
| 0.25x^2 | 0.5x+2 |
Since the geometric shapes have the same area, equating these expressions will produce a quadratic equation. 0.25x^2 = 0.5x+2 To write it in standard form, the Properties of Equality will be used.
This quadratic equation will now be solved by graphing. To do so, the quadratic function related to the equation will be written. Equation 0.25x^2 -0.5x - 2=0 ⇓ Related Function y=0.25x^2 -0.5x - 2 Next, the graph of the related function will be drawn. Identify its characteristics.
| y=0.25x^2 -0.5x - 2 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | y-intercept |
| a > 0 ⇓ upward | (1,- 2.25) | x=1 | (0,- 2) |
The reflection of the y-intercept across the axis of symmetry (2,- 2) is also on the parabola. With this information, the graph can be drawn.
From the graph, the x-intercepts of the function can be identified.
The parabola intersects the x-axis twice. The points of intersection are ( - 2,0) and ( 4,0). Therefore, the quadratic equation has two solutions, x_1= - 2 and x_2= 4. 0.25x^2 -0.5x - 2 =0 ↙ ↘ x_1 = - 2 x_2 = 4 Note that x cannot be negative because 0.5x represents the side length of the square. Therefore, although x=- 2 is a solution to the equation, only the solution x=4 makes sense in this context.
b The areas of the square and the triangle can be expressed as functions of x.
| Area of the Square | Area of the Triangle |
|---|---|
| f(x)= (0.5x)^2 | g(x)= 1/2 * 2 * (0.5x+2) |
| f(x) = 0.25x^2 | g(x)= 0.5x+2 |
To draw the graphs of these functions, a table of values will be made.
| x | f(x)=0.25x^2 | g(x)= 0.5x+2 |
|---|---|---|
| - 3 | 0.25( - 3)^2= 2.25 | 0.5( - 3)+2 = 0.5 |
| 0 | 0.25( 0)^2= 0 | 0.5( 0)+2 = 2 |
| 3 | 0.25( 3)^2= 2.25 | 0.5( 3)+2 = 3.5 |
Now, plot the points ( - 3, 2.25), ( 0, 0), and ( 3, 2.25) to graph f(x). Similarly, plot the points ( - 3, 0.5), ( 0, 2), and ( 3, 3.5) to graph g(x).
The graphs of the functions intersect at (- 2,1) and (4,4). Points of Intersection (- 2,1) and ( 4, 4) These points mean that the geometric shapes have the same area when x=- 2 or x=4. However, since x=- 2 makes the side length of the square negative, it should be discarded. Therefore, when x= 4, both figures have the same area, which is 4 square units.
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Example
Modeling Heights of Arrows With Quadratic Equations
Tiffaniqua and Ramsha are good at archery. They want to determine whether the arrows they shoot will collide in the air or not. Tiffaniqua takes her shot from a tree and Ramsha takes her shot from the ground just below Tiffaniqua. They wrote two quadratic functions to model the heights of the arrows in meters. Tiffaniqua:& y = - 1.2x^2+4.8x+8 Ramsha:& y = - 0.5(x-4)^2+8 In these equations, x represents the horizontal distance in meters from the point where the arrow is shot.
a Write a quadratic equation in standard form that will help determine whether the arrows will collide.
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b Find the answer that fits the context of the situation by solving the equation graphically.
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c Suppose a coordinate plane is placed on the diagram so that Ramsha is at the origin. How far is Ramsha from the point where the arrows collide? Write the answer in exact form.
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Hint
a At the collision point of the arrows, the functions have the same height. Therefore, equate the given quadratic functions.
b The equation written in the previous part has a related function. Draw the graph of this function.
c Find the coordinates of the point where the arrows collide. Then, use the Distance Formula.
Solution
a At the collision point of the arrows, if there is any, the functions have the same y-value. Therefore, to find this point, the right-hand sides of the equations can be equated.
y = - 1.2x^2+4.8x+8 y = - 0.5(x-4)^2+8 ⇓ - 1.2x^2 + 4.8x + 8 = - 0.5(x - 4)^2 + 8 This resulting quadratic equation needs to be written in standard form.
- 1.2x^2+4.8x+8 = - 0.5(x-4)^2+8
▼
Simplify right-hand side
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
- 1.2x^2+4.8x+8 = - 0.5(x^2-8x+16)+8
Distr
Distribute - 0.5
- 1.2x^2+4.8x+8 = - 0.5x^2+4x-8+8
SubTerm
Subtract term
- 1.2x^2+4.8x+8 = - 0.5x^2+4x
SubEqn
LHS-4x=RHS-4x
- 1.2x^2+0.8x+8 = - 0.5x^2
AddEqn
LHS+0.5x^2=RHS+0.5x^2
- 0.7x^2+0.8x+8 = 0
The solutions to this quadratic equation will give the horizontal distance to the collision point.
b The quadratic equation found in Part A will solved graphically. To do so, its related function will be drawn first.
Quadratic Equation - 0.7x^2+0.8x+8 = 0 ⇓ Related Function y = - 0.7x^2+0.8x+8 To be able to graph the function, some of its characteristics should be determined.
| y = - 0.7x^2+0.8x+8 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | y-intercept |
| a < 0 ⇓ downward | (4/7,288/35) | x=4/7 | (0,8) |
The reflection of the y-intercept across the axis of symmetry ( 87,8) is also on the parabola. Since these points are very close to each other, another point can be plotted by randomly choosing an x-value and calculating its corresponding y-value. Try x= - 3.
y = - 0.7x^2+0.8x+8
Substitute
x= - 3
y = - 0.7( - 3)^2+0.8( - 3)+8
▼
Evaluate right-hand side
CalcPow
Calculate power
y = - 0.7(9)+0.8(- 3)+8
Multiply
Multiply
y = - 6.3-2.4+8
AddSubTerms
Add and subtract terms
y= - 0.7
Therefore, the point (- 3, - 0.7) lies on the curve. The reflection of this point across the axis of symmetry is ( 297,- 0.7), which also lies on the parabola. With this information, the graph can be drawn.
The x-intercepts of the graph can now be identified.
The parabola intersects the x-axis twice. From the graph only one of the intercepts is identified easily, which is (4,0). The other intercept is between - 3 and - 2. However, this intercept can be ignored because x represents horizontal distance and distance cannot be negative. Therefore, in this context, only x=4 makes sense. Equation: & - 0.7x^2+0.8x+8 = 0 Solution: & x=4 This solution means that the horizontal distance from where Tiffaniqua and Ramsha stand to the point of collision is 4 meters.
c Place a coordinate plane so that Ramsha is at the origin. In Part B, the horizontal distance to the point where the arrows collide was found to be 4 meters
To find how far Ramsha is to the collision point, the Distance Formula will be used. d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) To do so, the y-coordinate of the collision point needs to be found. To accomplish this, substitute 4 into the either of the quadratic functions, and evaluate.
y = 0.5(x-4)^2+8
Substitute
x= 4
y = 0.5( 4-4)^2+8
▼
Evaluate right-hand side
DiffZero
a-a=0
y = 0.5(0)^2+8
CalcPow
Calculate power
y = 0.5(0)+8
ZeroPropMult
Zero Property of Multiplication
y=0+8
IdPropAdd
Identity Property of Addition
y=8
Therefore, the arrows collide at the point (4,8). Now the distance from Ramsha to the point where the arrows collide can be found.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstitutePoints
Substitute ( 0,0) & ( 4,8)
d = sqrt(( 4- 0)^2+( 8- 0)^2)
d = 4sqrt(5)
Ramsha is 4sqrt(5) meters away from the collision point of the arrows.
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Pop Quiz
Finding Solutions of Quadratic Equations
Find the solutions to each quadratic equation by graphing its related function. If there are two solutions, write the smaller solution first.
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Closure
Graphing Each Side of an Equation
In this lesson, real life situations that can be modeled by quadratic equations were solved graphically. Solving a quadratic equation graphically can be done in three steps. Step I: & Write the Equation in Standard Form Step II: & Graph the Related Function Step III: & Find thex-intercepts In general, when solving equations of form f(x)=g(x), the equation is rearranged and the graph of y=f(x)-g(x) is drawn. The zeros of the graph are solutions to the equation.
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Solving Quadratic Equations by Graphing
Exercise 3.1
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If the quadratic equation ax^2+bx+c=0 has no real solution, it means that the graph of its related quadratic function does not intercept the x-axis. Let's graph an arbitrary quadratic function with no real solution and whose vertex lies in the second quadrant.
Looking at the graph, we can see that the equation has no real solution if it opens upward. A parabola that opens upward has a positive leading coefficient. Therefore, a must be positive.
Let's translate the parabola so its vertex is in the fourth quadrant.
Moving the parabola to the fourth quadrant means that the graph lies below the x-axis. Because the graph opens upward, it intercepts the x-axis. Therefore, the graph has x-intercepts.
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