After firing the dummy from the cannon, the sensor recorded the dummy at a height of 416 feet at t = 1 (1 second after launch) and 704 feet at t = 10 (10 seconds after launch).

Question

1. At what height was the dummy at t = 0?

2. Using the 3 points, (0, 64), (1, 416) & (10, 704), derive a quadratic function h(t), that calculates the dummy’s

height at t seconds. Give the function in standard form h(t) = at2 + bt + c.

3. Give the function in vertex form h(t) = a(t − h)2 + k.

4. What is the maximum height the dummy will reach? SHOW WORK

5. How many seconds after launch did the dummy reach its maximum height?

6. What is the y-intercept?

7. What are the x-intercepts of this function?

8. What do the x and y intercepts represent in this problem?

9. How long (in seconds) did it take for the dummy to land?

10. Sketch the graph of this function labeling the key points and the axes appropriately.

Answer 1

Hi Richie E.

The y intercept is whenever t = 0 and is given

(0,64) so this answers 1, and 6

Use the three points and the Standard Form to set up three equations in 3 unknowns a, b, and c

f(x) = y = ax² + bx + c

h(t) = at² + bt + c

t = x and y = h(t)

Your three equations can be generated as follows:

(0,64)

64 = a(0)² + b(0) + c

64 = c

(1, 416)

416 = a(1)² + b(1) + c

416 = a + b + c

(10, 704)

704 = a(10)² + b(10) + c

704 = 100a + 10b + c

Since we already know that c = 64

We can set up two equations in two unknowns a and b

416 -64 = a + b

352 = a + b

704 - 64 = 100a + 10b

640 = 100a + 10b

Now we have

a + b = 352

100a + 10 b = 640

Multiply a + b = 352 by negative 10 then combine with 100a + 10b to eliminate b

-10a - 10b = -3520

100a + 10b = 640

90a = -2880

a = -32

Now we a = -32 and c = 64 if we substitute these in a + b + c = 416 we can solve for b

416 = -32 + b + 64

416 = 32 + b

416 - 32 = b

384 = b

Now we have all the coefficients for the Standard Form Equation

h(t) = -32t² + 384b + 64

We have a parabola that opens downward since a is negative

We can find the vertex form by finding the x coordinate of the vertex -b/2a = -384/2(-32) = -384/-64 = 6

We can find the y coordinate of the vertex by finding f(6)

f(6) = -32(6)² + 384(6) + 64 = 1216

Now just plug them in

f(t) = -32(x - 6)² + 1216

Since the parabola opens downward the vertex is coordinates of the highest point (6, 1216)

The maximum height is 1216 feet at t = 6 seconds

The x intercepts can be found by setting f(t) = 0 and using the Quadratic Formula to find t

(-0.164,0) and (12.16,0)

You have all that is needed to graph the function and answer the rest of the questions

Give the rest of it a try

You can graph this with a graphing calculator or at Desmos.com