When you throw a ball it first goes up into the air, slows as it travels, then stops and comes down again faster and faster.
See solution.
Practice makes perfect
Let's consider throwing a ball for this problem. When you throw a ball, first it goes up into the air — slowing as it travels — then it stops and comes down again faster and faster. We can model the height of the ball at all times using a quadratic function.
Let's add some more information to our example problem — keep in mind that this is just one of infinite possibilities we could use! We will assume that the ball is thrown straight up from 2 meters above the ground with a velocity of 9 m/s. Let t denote the time in seconds. We can work out the height by combining the following three things.
The ball starts at 2 m:
2
It travels upwards at 9 m/s:
9t
Gravity pulls it down with the rate about 5m per second squared:
- 5t^2
The height h at time t is the sum of the three expressions above.
h=3+9t-5t^2
We will determine how long the ball travels in the air. The ball will hit the ground when the height is zero, h= 0.
h=2+9t-5t^2 ⇓
0=2+9t-5t^2
The solutions, or roots, of at^2+bt+c=0 are the t-intercepts of the graph of h(t)=at^2+bt+c. Let's write our function in the standard form.
h(t)=2+9t-5t^2 ⇕ h(t)=-5 t^2+9t+3
Graphing the Function
To draw the graph of the function written in standard form, we must start by identifying the values of a, b, and c.
h(t)= - 5t^2+9t+2 ⇕ h(t)= - 5t^2+ 9t+ 2
We can see that a= - 5, b= 9, and c= 2. Now, we will follow three steps to graph the function.
The axis of symmetry is a vertical line with the equation of t=- 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 a vertical line with the equation t=0.9.
Making the Table of Values
Next, we will make a table of values using t-values around the axis of symmetry, t=0.9.
t
- 5t^2+9t+2
h(t)
- 1
- 5( - 1)^2+9( - 1)+2
- 12
0
- 5( 0)^2+9( 0)+2
2
1
- 5( 1)^2+9( 1)+2
6
2
- 5( 2)^2+9( 2)+2
0
3
- 5( 3)^2+9( 3)+2
- 16
Plotting and Connecting the Points
We can finally draw the graph of the function. Since a=- 5, which is negative, the parabola will open downwards. Let's connect the points with a smooth curve.
Finding the t-intercepts
Let's identify the t-intercepts of the graph of the related function.
We can see that the parabola intersects the t-axis twice. One point of intersection is negative. Therefore, it represents a negative time, which is impossible in our case. The second solution is t=2. This tells us that the ball hits the ground after 2 seconds!
