Question

an engineer uses the equation below to model the height of a cannonball shot straight up from ground level. in the model the height y (in feet) is a function of t, the number of seconds after the cannonball is shot.$y=-16t^{2}+32t$complete the parts below.(a) graph the parabola $y=-16t^{2}+32t$. to do so, plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph-a-function button.(b) write an interval below to write an inequality for all the values of t for which the cannonball is gaining height (going up). then choose the best description of this interval.inequality: $square < t < square$description:the function $y=-16t^{2}+32t$ is increasing for values of tthe function $y=-16t^{2}+32t$ is increasing past this interval, so

Explanation:

Step1: Find vertex of parabola

For $y=ax^2+bx+c$, vertex $x=-\frac{b}{2a}$.
Here $a=-16, b=32$, so $x=-\frac{32}{2(-16)}=1$.
Substitute $x=1$: $y=-16(1)^2+32(1)=16$.
Vertex: $(1,16)$

Step2: Pick left vertex points

Choose $x=0, x=-1$:

• $x=0$: $y=-16(0)^2+32(0)=0$ → $(0,0)$
• $x=-1$: $y=-16(-1)^2+32(-1)=-48$ → $(-1,-48)$

Step3: Pick right vertex points

Choose $x=2, x=3$:

• $x=2$: $y=-16(2)^2+32(2)=0$ → $(2,0)$
• $x=3$: $y=-16(3)^2+32(3)=-96$ → $(3,-96)$

Step4: Find increasing x-values

Parabola opens downward ($a<0$), so it increases from $x=0$ (launch) to vertex $x=1$.
Inequality: $0 < x < 1$

The function increases when $x$ is between 0 and 1, as the cannonball rises from ground to max height.

Answer:

(a) Points to plot:

$(-1,-48)$, $(0,0)$, $(1,16)$, $(2,0)$, $(3,-96)$ (plot these and draw the parabola)

(b)

Inequality: $0 < x < 1$
Description: The cannonball is rising (height increasing) from launch at $x=0$ until it reaches maximum height at $x=1$ second.