Question

Point symmetric to (t g) = 4 0). (t g) =

Answer 1

this is a quadratic function, therefore it has the form of a parabola.

We can find the cut-off points with the x-axis and the y-axis by making t = 0 and y = 0 , respectively

when t = 0 , then g(t) = g(0) = -0.5*0 + 8 = 8 , Therefore, the cut with the y axis is at the point (0,8)

when g(t) = 0 , then 0 = -0.5t^2 + 8

`\begin{gathered} 0.5t^2=8 \\ t^2=(8)/(0.5)=(8)/((1)/(2))=8\cdot2=16 \\ t=\sqrt[]{16} \\ t=\pm4 \end{gathered}`

therefore, the cut-off with the x axis are ( -4 , 0) and ( 4 , 0)