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| 14 Theory slides |
| 11 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.
Consider four different graphs of functions. Examine them closely to determine both similarities and differences.
A quadratic expression is a specific type of polynomial that follows the form ax2+bx+c. Each term has a specific name.
For an expression in this form to be quadratic, a must be a non-zero number. Consider a few examples of quadratic functions.
Quadratic Expression |
---|
x2+3x+5 |
5x2+41x+23 |
53x2+21x+4.3 |
Commutative Property of Addition
Add and subtract terms
Commutative Property of Addition
6x3−6x3=0
Identity Property of Addition
Commutative Property of Addition
Add and subtract terms
Commutative Property of Addition
Consider the given algebraic expression. Determine whether it is a quadratic expression by combining like terms.
A parabola can be vertical or horizontal. A vertical parabola can open upward or downward. In comparison, a horizontal parabola can open to the left or the right. The graph of a quadratic function is a vertical parabola.
A parabola either opens upward or downward. This is the direction of the parabola . If the leading coefficient a of the corresponding equation is positive, the parabola opens upward. If the coefficient is negative, the parabola opens downward.
A quadratic function is a polynomial function of degree 2 that can always be written in the form y=ax2+bx+c, with a=0. Notice that the highest exponent of the independent variable is 2. The graph of any quadratic function is a vertical parabola.
Consider the graph of a function. Identify whether the graph shows a parabola or not.
Consider the graph of a quadratic function. Is the leading coefficient of the parabola positive or negative?
Each pair of numbers in the table of values is an ordered pair that represents the coordinates of a point on the graph. Plot all the points and then connect them with a smooth curve.
The following table of values is given for the bridge in the shape of a parabola. There, x represents the distance from the base of one side of the bridge, while y represent the height of the bridge over the ground.
x | y |
---|---|
0 | -0.6 |
1 | 0.5 |
2 | 1.4 |
4 | 2.6 |
6 | 3 |
8 | 2.6 |
10 | 1.4 |
12 | -0.6 |
Note that each pair of values of x and y forms an ordered pair (x,y). These ordered pairs are coordinates of the points on the parabola. Start by plotting them on a coordinate plane.
Now, connect the points with a smooth curve. This way the graph of the parabola is drawn.
The crew also brought some species of animals from the Earth. They started observing the changes in their populations over time. Here is the table of values that shows the changes in kangaroo population over the period of one year.
Use the table of values to graph the population growth of kangaroos. Note that x represents the number of month since the observation began and y represents the population of kangaroos in the thousands.
Each pair of numbers in the table of values is an ordered pair that represents the coordinates of a point on the graph. Plot all the points and then connect them with a smooth curve.
The following table of values represents the population of kangaroos over one year. Here, x represents the number of month since the observation began and y represents the population of kangaroos in thousands.
x | y |
---|---|
0 | 30 |
2 | 20 |
4 | 14 |
6 | 12 |
8 | 14 |
10 | 20 |
12 | 30 |
Note that each pair of values of x and y forms an ordered pair (x,y). These ordered pairs are coordinates of the points on the parabola. Start by plotting them on a coordinate plane.
Now, connect the points with a smooth curve. This way the graph of the parabola is obtained.
Each pair of numbers in the table of values is an ordered pair that represents the coordinates of a point on the graph. Plot all the points and then connect them with a smooth curve.
Consider the table of values that represents the total number of communication devices y on Harmonica after x days of working on this task.
x | y |
---|---|
0 | 0 |
1 | 0.8 |
2 | 3.2 |
3 | 7.2 |
4 | 12.8 |
5 | 20 |
6 | 28.8 |
Note that each pair of values of x and y forms an ordered pair (x,y). These ordered pairs are coordinates of the points on the parabola. Start by plotting them on a coordinate plane.
Now, connect the points with a smooth curve. This way the graph of the parabola is obtained.
Finally, use the graph to determine how many communication devices there will be on Harmonica on Day 7. Draw an arrow from x=7 to the parabola and then from the parabola to the y-axis.
Therefore, there will be about 40 communication devices on Day 7.
Maya really enjoys playing tennis. When she hits a ball, the ball leaves the racket 0.6 meters above the ground. The equation h=-4.9t2+4t+0.6 gives the ball's height h in meters after t seconds.
We know the equation that represents the ball's height h in meters t seconds after Maya hits the ball. h=- 4.9t^2+4t+0.6 We are asked to determine the time when the ball will be at the highest point. Let's start with making a table of values. We will substitute different values of t into the equation and evaluate it for h.
t | h=- 4.9t^2+4t+0.6 | h |
---|---|---|
0 | - 4.9( 0)^2+4( 0)+0.6 | 0.6 |
0.2 | - 4.9( 0.2)^2+4( 0.2)+0.6 | ≈ 1.2 |
0.4 | - 4.9( 0.4)^2+4( 0.4)+0.6 | ≈ 1.4 |
0.6 | - 4.9( 0.4)^2+4( 0.4)+0.6 | ≈ 1.2 |
0.8 | - 4.9( 0.4)^2+4( 0.4)+0.6 | ≈ 0.7 |
0.9 | - 4.9( 0.4)^2+4( 0.4)+0.6 | ≈ 0.2 |
This way we got ordered pairs (t,h) that are the coordinates of the points on the graph. Let's plot them!
Next, we can connect the points to get the graph of the equation.
We can see that the ball reaches the highest point in its path at the vertex of the parabola. Its coordinates are (0.4,1.4). This means that the ball reaches its highest point 0.4 seconds after Maya hits it.
Recall Maya hits the ball and it leaves the racket at 0.6 meters above the ground. This is why, at t=0 the value for the function modeling the height is 0.6.
Therefore, at t=0 the height is 0.6 meters. From Part A, we know that at t=0.4 seconds the ball is at its maximum height. After another 0.4 seconds, when t=0.8 seconds, the ball will again be at about 0.6 meters, so not on the ground! This is due to the symmetry of the parabola.