1) Consider y = f(x) where f(x) = x²+4x-12
[Consult photo attachment 1 for a graph of f(x) = x²+4x-12]
a) Find the x-intercepts:
To find the x-intercepts, we must find the values of x which satisfy f(x) = 0
x²+4x-12 = 0
(x+6)(x-2) = 0
x = 2, x = -6
The x-intercepts of f(x) are (2, 0) and (-6, 0)
b) Find the y-intercept:
To find the y-intercept, find the value of f(x) at x = 0
f(0) = 0²+4×0-12 = -12
The y-intercept of f(x) is (0, -12)
c) What is the maximum or minimum value of f(x)?
We can rewrite f(x) by completing the square.
f(x) = x²+4x-12
f(x) = (x²+4x+4) -12-4
f(x) = (x+2)²-16
You can think of f(x) as the result of shifting the graph y = x² to the left 2 units and down 16 units.
We can now tell from this that f(x) is at its minimum value when x+2 = 0
x = -2, now plug this into f(x)
f(-2) = (-2+2)²-16 = -16
The minimum value of f(x) is -16
2) Consider the graphs y = x²+3x-5 and y = 4x+1
To find where these graphs intersect, find the values of x for which they are equal to each other.
x²+3x-5 = 4x+1
x²-x-6 = 0
(x-3)(x+2) = 0
x = 3, x = -2
Now that we have the x-coordinates of the intersection points, plug these x values into either of the two graphs to find the y-coordinates of the intersection points.
Plugging the x values into y = 4x+1 is less work.
y = 4(3)+1 = 13
y = 4(-2)+1 = -7
The intersection points are (3, 13) and (-2, -7)
Seth says the intersection points are (-2, 0) and (3, 0). Seth's solution is incorrect because the y-coordinates of the intersection points are not 0. However, he shows the correct work in determining the x-coordinates of the intersection points. He may have forgotten to plug in the x-values to get the y-coordinates of the intersection points and instead just assumed the y-coordinates of both points were 0.
3) A rocket is launched from the ground with an initial velocity of 64ft/s.
d) Write a quadratic function h(t) that shows the height, in feet, of the rocket t seconds after it was launched.
Apply this kinematics equation from physics:
d(t) = d₀ + v₀t + 0.5at²
Change the variable to h to represent height as a function of time t:
h(t) = h₀ + v₀t + 0.5at²
h₀ is the rocket's initial height, 0 ft (launched from the ground)
v₀ is the rocket's initial upward velocity, 64ft/s
a is the downward acceleration of the rocket due to earth's gravity, -32ft/s²
h(t) = 0 + 64t + 0.5(-32)t²
h(t) = -16t² + 64t
e) Graph h(t)
[Consult photo attachment 2 for a graph of h(t) = -16t²+64t]
f) Use the graph to determine the rocket's maximum height, the amount of time it took to reach its maximum height, and the amount of time it was in the air.
It is easy to look at the graph to determine these answers:
- The rocket's maximum height is 64 feet.
- The amount of time the rocket takes to reach its maximum height is 2 seconds.
- The amount of time the rocket was in the air is 4 seconds.
But just in case you want to solve this question without looking at the graph:
- The formula for finding the x-coordinate of the vertex of a quadratic function of the form f(x) = ax²+bx+c is x = -b/2a. h(t) = -16t²+64t... t = -64/2(-16) = 2. h(2) = 64, so the maximum height is 64 feet.
- We already found the time it takes to reach the rocket's maximum height in the work to find the maximum height, 2 seconds.
- To find how long the rocket was in the air, find the time t where the height of the rocket is 0 feet, which means find values of t which satisfy h(t) = 0... -16t²+64t = 0... 16t²-64t = 0... 16(t²-4t) = 0... 16(t(t-4)) = 0 t = 4 and t = 0, but reject t = 0 because we already know the rocket starts on the ground at t = 0... so the rocket was in the air for 4 seconds.
