Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 8.2
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Exercise 109 Page 406

Practice makes perfect
a In any function, the constant shows the graph's y-intercept. Therefore, by examining the function, we can identify the y-intercept as 11.27.

h=-4.9t^2+49t+ 11.27 ← constant This means that the platform was 11.27 feet above ground.

To find the vertex, we should rewrite the function into graphing form. graphing form:& y=a(x- h)^2+ k vertex:& ( h, k) To do that, we have to complete the square. However, this requires the squared variable to have a coefficient of 1. Therefore, we will first divide both sides of the equation by - 4.9. h=- 4.9t^2+49t+11.27 ⇓ h/- 4.9=t^2-10t-2.3 Now we can complete the square and then write the function in graphing form.

h/- 4.9=t^2-10t-2.3
h/- 4.9+(- 10/2)^2=t^2-10t-2.3+(- 10/2)^2
Solve for h
h/- 4.9+(10/2)^2=t^2-10t-2.3+(10/2)^2
h/- 4.9+5^2=t^2-10t-2.3+5^2
h/- 4.9+5^2=t^2-10t+5^2-2.3
h/- 4.9+5^2=t^2-2(t)(5)+5^2-2.3
h/- 4.9+5^2=(t-5)^2-2.3
h/- 4.9+25=(t-5)^2-2.3
h/- 4.9=(t-5)^2-27.3
h=- 4.9(t-5)^2+133.77

Having rewritten the equation in graphing form, we can identify its vertex to (5,133.77). Let's add this point to the coordinate plane.

The number of seconds that pass before the fireworks hits the ground is given by the graph's second root. To calculate this, we should equate the graphing form of the equation by 0 and solve for t.

h=- 4.9(t-5)^2+133.77
0=- 4.9(t-5)^2+133.77
Solve for t
- 133.77=- 4.9(t-5)^2
27.3=(t-5)^2
(t-5)^2=27.3
t-5=± sqrt(27.3)
t=5± sqrt(27.3)
lct=5+sqrt(27.3) & (I) t=5-sqrt(27.3) & (II)

(I), (II): Use a calculator

lt=10.22494... t=- 0.224940...

(I), (II): Round to 2 decimal place(s)

lt_1≈ 10.22 t_2≈- 0.22

The function has two roots at t≈ - 0.22 and t≈ 10.22. Since the function measures the time passed since the firework was launched at t=0, we can discard the negative root. The positive root tells us that after 10.22 seconds the firework hits the ground. With this information, we can graph the function.

Let's summarize what we have found. Height of platform:& 11.27 feet Maximum height:& 133.77 feet Seconds until h=0:& 10.22 seconds

b To write the equation in factored form, we have to know its roots. In Part A we calculated the function's roots. t_1=5-sqrt(27.3) t_2=5+sqrt(27.3) By substituting these zeroes into the following equation, we can write the function in factored form.
h=a(t-b)(t-c) In this form b and c are the function's zeroes and a is the stretch factor. Note that the stretch factor is the same as the coefficient to the variable with the highest degree — in this case, t^2. With this information, we can write the equation in factored form. h=- 4.9(t-(5-sqrt(27.3)))(t-(5+sqrt(27.3))) In this form we can immediately identify the equation's zeroes.