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| 9 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
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.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.
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.
Now, find all the points on the graph that have a y-coordinate equal to 0.
The x-coordinates of the identified points solve the equation f(x)=0.
x= - 2
Calculate power
Multiply
a(- b)=- a * b
a-(- b)=a+b
Add and subtract terms
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
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.
For each quadratic equation, draw its related quadratic function to determine the number of solutions.
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.
LHS-192=RHS-192
Rearrange equation
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.LHS-196=RHS-196
Rearrange equation
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.
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.
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 |
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.
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 |
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-b)^2=a^2-2ab+b^2
Distribute - 0.5
Subtract term
LHS-4x=RHS-4x
LHS+0.5x^2=RHS+0.5x^2
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) |
x= - 3
Calculate power
Multiply
Add and subtract terms
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.
x= 4
a-a=0
Calculate power
Zero Property of Multiplication
Identity Property of Addition
Substitute ( 0,0) & ( 4,8)
Find the solutions to each quadratic equation by graphing its related function. If there are two solutions, write the smaller solution first.
Use the graph to solve the quadratic equation. When entering the answer, use the labels to remove or add to the number of roots being entered.
Equation: x^2 +8x + 7 = 0
Equation: - x^2 -8x-12 = 0
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 .
Solve the equation by graphing.
We are asked to solve the given quadratic equation by graphing. To do so, these three steps can be followed.
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
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.
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.
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).
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.
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.
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.
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
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.
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.
Magdalena's kindergarten-aged little brother is learning to play golf.
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.
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.
The equation models the height h of a soccer ball in centimeters after Maya kicks it, where t is the time measured in seconds. h = - 4t^2+40t+44
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.
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
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.
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.
Consider the graph shown.
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.
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.