When a quadratic equation is in standard form, what does the constant term represent? What is the t-coordinate at this point?
B
b
When the coefficient of the quadratic term is negative, where does the parabola open? What will be the direction in which the object is moving?
C
c
Equate the function h(t) and the height you are given. Solve this equation for t.
D
d
Equate the function h(t) and the height you are given. Solve this equation for t.
E
e
Make a graph of the given function. Can t be negative? Can h(t) be negative? What is an appropriate domain?
A
a
The height of the object above the ground at time t=0 seconds is 1700 feet.
B
b
Since the coefficient is negative, the object is moving downwards.
C
c
t=6.614 seconds
D
d
t=6.892 seconds
E
e
Example Domain: From 0 to 10.4
Example Range: From 0 to 1700
Practice makes perfect
a
We are given the function below which models the height h of an object (in feet) at t seconds.
h(t) = -16t^2+ 1700
In a quadratic function in standard form y=ax^2+bx+c, we have that the constant c represents the y-intercept. In our case, the h-intercept is 1700. Also, at the h-intercept the t-coordinate is 0, which means that the height of the object at t=0 seconds is 1700 feet.
b
Since the coefficient of t^2 is negative ( -16), it implies that the graph of the equation is a parabola that opens downward. Now, let's find the vertex of this parabola knowing that a= -16 and b=0.
Vertex
x-coordinate
y-coordinate
x=-b/2a
y=h(-b/2a)
x=-0/2(-16)=0
y=h(0)=1700
The vertex of our parabola is (0,1700), and because the parabola opens downward, we can conclude that the maximum height of the object is 1700 feet. Therefore, the direction in which the object is moving is downward. Let's illustrate with a graph.
c
To know when the object will be 1000 feet above the ground, we need to solve the equation h(t)=1000.
h(t) = 1000 ⇒ -16t^2 + 1700 &= 1000
-16t^2 + 700 &= 0
To solve the equation above, we will use a graphing calculator. We begin by pushing the Y= button and typing the equation in the first row. Next, we push GRAPH to draw it.
For this function, we cannot see the entire graph in the standard viewing window. We can resize our window by pushing WINDOW, changing the settings, and once more pushing GRAPH. Note that the constant term is 700, so this is the y-intercept. Recognizing this gives us an idea of how high our window should go.
Next, to find the zero, push 2nd and TRACE. From this menu, choose zero. To help the calculator work faster, we will have to provide it with some guesses as to where the zero is. To make sure the calculator picks the zero we want to find, we will make our guess as close as possible to it.
In conclusion, the object will be 1000 feet above the ground at t=6.614 seconds.
d
This time, we have to find when the object will be 940 feet above the ground. As in Part C, to find the answer, we will need to solve the equation h(t)=940.
lc t ≥ 0:t = 6.892 & (I) t < 0:t = - 6.892 & (II)
lt=6.892 t=-6.892
Since the variable t represents time, we discard the negative option. Consequently, the object will be 940 feet above the ground at t=6.892 seconds.
e
The function h(t) = -16t^2+1700 represents the height of an object h at t seconds. Since t measures the time it cannot be negative. Also, the height above ground can never be negative. To find an appropriate domain, let's take a look at the graph we made in Part B.
From the graph we conclude an appropriate domain for h(t) is 0≤ t ≤ 10.4, and its range will be 0 ≤ h(t) ≤ 1700.
