is the t(t-2)^2 in the denominator? Or only the 4? I can't tell when you write it like that.
Since your independent variable is t, I assume you want t-intercepts. To get t intercepts (where y = 0), set the equation equal to zero and solve for the t values that are roots.
I think the vertex from 1/4t(t-2)^2 is -2,0. But I'm not sure how to get the other vertices, as well as the x intercepts. Can you explain, please? THanks!
2 answers
It's (1/4)t...