The equation showing the path of the donut is given to be:
[tex]y=-x^2+4x-2[/tex]
QUESTION A
We can use the points where x = 1, 2, 3.
We have that:
[tex]\begin{gathered} When\text{ }x=1,y=-(1)^2+4(1)-2=1 \\ When\text{ }x=2,y=-(2)^2+4(2)-2=2 \\ When\text{ }x=3,y=-(3)^2+4(3)-2=1 \end{gathered}[/tex]
The table is shown below:
The graph is shown below:
QUESTION B
The maximum height can be seen in the graph.
From the graph, the maximum height is at the point:
[tex]x=2[/tex]
QUESTION C
The maximum height is at the point:
[tex]y=2[/tex]
QUESTION D
The vertex form of a quadratic equation is given to be:
[tex]y=a(x-h)^2+k[/tex]
The vertex of the graph is:
[tex](h,k)=(2,2)[/tex]
Therefore, we have:
[tex]y=a(x-2)^2+2[/tex]
At the point:
[tex](x,y)=(1,1)[/tex]
we can calculate a to be:
[tex]\begin{gathered} 1=a(1-2)^2+2 \\ 1=a+2 \\ a=1-2 \\ a=-1 \end{gathered}[/tex]
Therefore, the equation will be:
[tex]y=-(x-2)^2+2[/tex]
