a Start with finding the axis of symmetry. Then, plot the vertex and two other points.
B
b Look for the h-intercept, use t=0.
C
c Use the h-value of the vertex from Part A.
D
d Use the mirrored paired point from the h-intercept from Part A.
E
e A reasonable domain, t-values, and range, h-values, is when and where the ball is above the ground.
A
a
When h > 0 the ball is above ground. Both as t approaches negative infinity and positive infinity h goes to negative infinity. Once the ball reaches a height of 0, the ball would be represented as below ground.
B
b The ball is hit from the ground, h=0.
C
c 50 m
D
d 6.4 sec
E
eRange: 0 ≤ h ≤ 50 Domain: 0 ≤ t ≤ 6.4
Practice makes perfect
a To graph a quadratic function, we need to find points in the following way.
Find the axis of symmetry.
Find the vertex.
Find the h-intercept.
Find the corresponding point from the h-intercept across the axis of symmetry.
Connect the points in the shape of a parabola.
Let's start with the axis of symmetry. In the function h=-4.9t^2+31.3t, a=-4.9
and b=31.3.
We need two more points. Let's start with the h-intercept, then find it's mirrored corresponding point. To find the h-intercept, we need to substitute t=0 into the function.
The point (0,0) can be mirrored across the axis of symmetry by keeping the h-value and doubling the distance from the t-value to the vertex. The distace from 0 to the axis of symmetry is 3.2. When we double that we get 6.4, so our point is (6.4,0). Let's plot these points on the graph.
Now, let's connect the points in the shape of a parabola.
b To find the starting point of the ball, we need to evaluate the function at t=0. We did this in Part A. We found (0,0) to be the h-intercept. Therefore, the ball starts at ground level 0 meters above the ground.
c The maximum height of a function is the h-value of the vertex. In Part A, we found the vertex at (3.2,50). The h-value is 50 meters off the ground and thus the maximum height.
d The length of time it took the ball to hit the ground can be found from the t-value of the function when h=0. We found this point in Part A when we mirrored the h-intercept and got (6.4,0) Therefore, it took the ball 6.4 seconds to hit the ground.
e Given that we are plotting a function that represents the height of a ball over time, only non-negative h-values are reasonable. If we look at the graph, we can see that the function is non-negative over the t-values between 0 and 6.4 and the h-values between 0 and about 50.
