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
Exercises
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To determine the solutions of the quadratic equation, we need to know the graph of its related function. Quadratic Equation x^2+8x+7=0 ⇓ Related Function y = x^2+8x+7 The solutions of the equation are the x-coordinates of the zeros of function. Since we are given the graph of the related function, we can find them.
The graph crosses the x-axis twice, so the equation x^2+8x+7=0 has two solutions. We can see that those solutions are x=- 7 and x=- 1.
We want to use the given graph to determine the solutions of the given quadratic equation. Recall that the solutions of a quadratic equation are the zeros of its related function.
Quadratic Equation
- x^2-8x-12=0
⇓
Related Function
y = - x^2-8x-12
We are given the graph of the related function.
This graph crosses the x-axis twice, so its related equation - x^2+2x+3=0 has two solutions. We can see that those solutions are x=- 2 and x=-6 .
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Hint & Answer
S
Solution
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We are asked to solve the given quadratic equation by graphing. To do so, these three steps can be followed.
- Write the equation in standard form, ax^2+bx+c=0.
- Graph the related function y=ax^2+bx+c.
- Find the x-intercepts, if any.
The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. Let's identify its related function. Equation:&x^2-4x+4=0 Related Function:&y=x^2-4x+4
Graphing the Related Function
To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=x^2-4x+4 ⇓ y= 1x^2+( -4)x+ 4 We can see that a= 1, b= -4, and c= 4. Now, we will follow four steps to graph the function.
- Find the axis of symmetry.
- Calculate the vertex.
- Identify the y-intercept and its reflection across the axis of symmetry.
- Connect the points with a parabola.
Finding the Axis of Symmetry
The axis of symmetry is a vertical line with equation x=- b2 a. Since we already know the values of a and b, we can substitute them into the formula.
The axis of symmetry of the parabola is the vertical line with equation x=2.
Calculating the Vertex
To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2 a, f( - b/2 a ) ) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=2. Therefore, the x-coordinate of the vertex is also 2. To find the y-coordinate, we need to substitute 2 for x in the given equation.
We found the y-coordinate, and now we know that the vertex is (2,0).
Identifying the y-intercept and its Reflection
The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Therefore, the point where our graph intercepts the y-axis is (0, 4). Let's plot this point and its reflection across the axis of symmetry.
Connecting the Points
We can now draw the graph of the function. Since a= 1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.
Finding the x-intercepts
Let's identify the x-intercepts of the graph of the related function.
We can see that the parabola intersects the x-axis once. The point of intersection is ( 2,0). Therefore, the equation x^2-4x+4=0 has one solution, x= 2.
We can check our solution by substituting the value into the given equation. If our solution is correct, the final result will show 0 on both sides of the equation. Let’s check x = 2.
We will follow the same steps to solve the given quadratic equation.
- Graph the related function y=ax^2+bx+c.
- Find the x-intercepts, if any.
The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Let's first write our equation in the standard form.
Now that we have our equation written in the standard form, let's identify the function related to the equation. Equation:&x^2+8x+28=0 Related Function:&y=x^2+8x+28
Graphing the Related Function
To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. y=x^2+8x+28 ⇓ y= 1x^2+ 8x+ 28 We can see that a= 1, b= 8, and c= 28. Now we can identify the characteristics of the function.
| y= x^2+8x+28 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | y-intercept |
| a > 0 ⇓ upward | (- 4,12) | x=- 4 | (0,28) |
The reflection of the y-intercept across the axis of symmetry (- 8,28) is also on the parabola. With this information, we can draw the graph.
Finding the x-intercepts
Let's identify the x-intercepts of the graph of the related function.
We can see that the parabola does not intersect the x-axis. Therefore, the equation x^2+8x+28=0 has no solution. This also means that the original equation, - x^2 = 8x+28 has no solution.
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Magdalena's kindergarten-aged little brother is learning to play golf.
The given equation models the height y in meters of his golf shot. y= - x^2+6x In this equation, x is the horizontal distance in meters. How far away does the golf ball land?
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We are given an equation that models the height y of a golf shot depending on the horizontal distance x. y=- x^2+6x Because the ball lands when its height is 0 meters, we can substitute 0 for y in our equation. 0=- x^2+6x We will solve this quadratic equation by graphing. There are three steps for solving a quadratic equation by graphing.
- Write the equation in standard form, ax^2+bx+c=0.
- Graph the related function y=ax^2+bx+c.
- Find the x-intercepts of the graph, if any.
The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. We only have to rearrange it! 0=- x^2+6x ⇔ - x^2+6x=0 Let's write the function related to our equation. Equation:&- x^2+6x=0 Related Function:& y=- x^2+6x To graph the function y=- x^2+6x, we begin by determining the values of a, b, and c. y=- x^2+6x ⇓ y= - 1x^2+ 6x+ 0 Therefore, a= - 1, b= 5, and c= 0. Then we identify the axis of symmetry.
The axis of symmetry for this function is x=3. Next, we determine its vertex. The axis of symmetry always intersects the parabola at its vertex. Therefore, x=3 is the x-coordinate of the vertex. We can use this to determine the corresponding y-coordinate.
The vertex of the function is (3,9). Recall that c= 0, so the y-intercept is (0, 0). Finally we plot the axis of symmetry x=3, the vertex (3,9), and the y-intercept (0,0). We also reflect the y-intercept across the axis of symmetry.
Finally, we connect the three points with a smooth curve.
The x-intercepts of the graph are x=0 and x=6. However, x=0 represents the distance of the ball at the beginning of the shot. Therefore, x=6 is the only solution that makes sense in this context. This means that the golf ball lands 6 meters away from the golfer.
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We are given an equation that models the height h of a soccer ball depending on the time t. h=- 4t^2+40t+44 We want to determine after how many seconds the ball hits the ground. The ball lands when its height is 0 feet. Therefore, h= 0. 0=- 4t^2+40t+44 We will solve this quadratic equation by graphing. There are three steps to follow when solving a quadratic equation using the graphing method.
- Write the equation in standard form, ax^2+bx+c=0.
- Graph the related function y=ax^2+bx+c.
- Find the x-intercepts, if any.
The solutions of ax^2+bx+c=0 are the x-intercepts of the graph of y=ax^2+bx+c. Our equation is already written in standard form. We only have to rearrange it! 0=- 4t^2+40t+44 ⇓ - 4t^2+40t+44=0 Let's write the related function of our equation. Equation:&- 4t^2+40t+44=0 Related Function:&h(t)=- 4t^2+40t+44
Graphing the Related Function
To draw the graph of the related function written in standard form, we must start by identifying the values of a, b, and c. h(t)=- 4t^2+40t+44 ⇓ h(t)= - 4t^2+ 40t+ 44 We can see that a= - 4, b= 40, and c= 44. Now we can identify the characteristics of the function.
| h(t)= - 4t^2+40t+44 | |||
|---|---|---|---|
| Direction | Vertex (- b/2a, h(- b/2a )) | Axis of Symmetry t = - b/2a | h -intercept (0,c) |
| a < 0 ⇓ downward | (5,144) | t = 5 | (0,44) |
The reflection of the h-intercept across the axis of symmetry, (10,44), is also on the parabola. With this information, we can draw the graph.
Finding the t-intercepts
We can see that the graph has two t-intercepts, one negative and one positive.
Recall that t represents the time after the ball is kicked. Therefore, it cannot be negative. Only the positive t-intercept has meaning in this situation. The positive t-intercept is t=11. This means that the ball hits the ground after 11 seconds.
To find the time when the ball is at a height of 108 centimeters, we need to substitute 108 for h into the given quadratic equation. Then, we will write it in standard form.
Now we can write the function related to the equation. Equation: & - 4t^2 + 40t -64 = 0 Related Function: & h(t)=- 4t^2 +40t - 64 We can now identify the characteristics of the function.
| h(t)=- 4t^2 + 40t - 64 | |||
|---|---|---|---|
| Direction | Vertex | Axis of Symmetry | h-intercept |
| a < 0 ⇓ downward | (5,36) | t= 5 | (0,- 64) |
In addition to the vertex and the h-intercept, the reflection of the h-intercept across the axis of symmetry, which is at (10,- 64), is also on the parabola. With this information, the graph of the function can be drawn.
The parabola intersects the t-axis twice. The points of intersection are ( 2,0) and ( 8,0).
Therefore, the equation - 4t^2 + 40t-64=0 has two solutions, t_1= 2 and t_2= 8.
- 16t^2 +112t-192=0
↙ ↘
t_1 = 2 t_2 = 8
This means that the ball reaches the height of 108 centimeters twice. After 2 seconds and again after 8 seconds.
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Usually, to solve a quadratic equation by graphing, we rewrite the equation in standard form, graph its related function, and determine its x-intercepts. However, we can also graph each side of the equation. In this case, the solutions of the equation are the x-coordinates of the points of intersection of the graphs. Let's consider the given equation. x^2= - 2x+8 ⇒ lLHS:y= x^2 RHS:y= - 2x+8 The function related to the left-hand side of the equation is y=x^2, and the function related to the right-hand side of the equation is y=- 2x+8. Note that these functions are plotted on the given graph.
Recall that the solutions of the equation are the x-coordinates of the points of intersection of the graphs related to the sides of the equation. Let's identify them!
The graphs intersect twice. Therefore, the equation x^2=- 2x+8 has two solutions. These solutions are x=- 4 and x=2.
We need to describe what we know about the graph of y=x^2+2x-8. Let's start by identifying the values of a, b, and c. These values can tell us a lot about the graph of a quadratic function. y=x^2+2x-8 ⇓ y= 1x^2+ 2x+( - 8) We have a= 1, b= 2, and c= - 8. First of all, since a>0, we know that the parabola opens upwards. Next we will use the values of a and b to identify the axis of symmetry.
The axis of symmetry for this function is x=- 1. Next we can determine the vertex of the parabola. The axis of symmetry always intersects the parabola at its vertex. Therefore, x=- 1 is the x-coordinate of the vertex. We will use this to determine the corresponding y-coordinate.
The vertex of the parabola is (- 1,- 9). Finally recall that c is the y-intercept of the parabola. Therefore, - 8 is the y-intercept of our parabola. This is all that we can tell about the graph based on the values of a, b, and c. Now we will compare the function from Part B with the equation from Part A. Equation:& x^2=- 2x+8 Function:& y=x^2+2x-8 Let's rewrite the equation in standard form.
We can see that the function related to the equation from Part A is the function from Part B! Equation: & x^2+2x-8 = 0 Related Function: & y = x^2+2x-8
The solutions of the equation are the x-intercepts of the parabola. From Part A we know that the equation has two solutions. Thus, the related function must have two x-intercepts. These x-intercepts are x=- 4 and x=1. Let's sum up the information about the graph of y=x^2+2x-8.
- The parabola faces upwards.
- The axis of symmetry of the parabola is the vertical line x=- 1.
- The vertex is the point (- 1,- 9).
- The y-intercept of the parabola is y=- 8.
- The x-intercepts of the parabola are x=- 4 and x=2.
Therefore, the statements in I and II are correct. To confirm our observations, we have included a graph of the function y=x^2+2x-8.
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Solving Quadratic Equations by Graphing
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