Question

1. a cannonball is shot into the air. its height during its flight can be described by the equation h = -5(t - 1)(t - 9) where h is height in metres and t is time in seconds.

(a) what are the zeroes of this relation?
(b) what do the zeroes mean in this situation?
(c) what is the axis of symmetry and what does it represent?
(d) use the axis of symmetry to find the maximum height of the cannonball.

Explanation:

Step1: Find zero - es of the relation

Set \(h = 0\), so \(-5(t - 1)(t - 9)=0\). Then \(t-1 = 0\) or \(t - 9=0\).

Step2: Solve for \(t\)

If \(t-1 = 0\), then \(t = 1\); if \(t - 9=0\), then \(t=9\).

Step3: Interpret zero - es in context

The zero - es \(t = 1\) and \(t = 9\) mean the times when the cannonball is at ground level (\(h = 0\)). \(t = 1\) could be the time when the cannonball is fired (or passes a certain reference point on the way up) and \(t = 9\) is the time when it lands.

Step4: Find axis of symmetry

The formula for the axis of symmetry of a quadratic function in factored form \(y=a(x - r_1)(x - r_2)\) is \(x=\frac{r_1 + r_2}{2}\). Here \(r_1 = 1\), \(r_2 = 9\), so the axis of symmetry is \(t=\frac{1 + 9}{2}=5\). It represents the time at which the cannonball reaches its maximum height.

Step5: Find maximum height

Substitute \(t = 5\) into the height equation \(h=-5(t - 1)(t - 9)\). Then \(h=-5(5 - 1)(5 - 9)=-5\times4\times(-4)=80\).

Answer:

(a) \(t = 1\) and \(t = 9\)
(b) The times when the cannonball is at ground - level (\(h = 0\)).
(c) The axis of symmetry is \(t = 5\). It represents the time when the cannonball reaches its maximum height.
(d) The maximum height of the cannonball is \(80\) metres.