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!
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.
It's (1/4)t...
Certainly! To find the vertex of the quadratic equation given by 1/4t(t-2)^2, you can use the formula x = -b/2a, where the equation is in the form ax^2 + bx + c. In this case, a = 1/4 and b = -1.
First, let's find the x-coordinate of the vertex. Plugging these values into the formula, we have:
x = -(-1) / 2(1/4)
x = 1 / (1/2)
x = 2
So the x-coordinate of the vertex is 2.
To find the y-coordinate of the vertex, you need to substitute this x-coordinate back into the original equation. When t = 2:
y = 1/4(2)(2-2)^2
y = 1/4(2)(0)^2
y = 0
Therefore, the vertex is (2, 0).
To find the other vertices, you can plug in different values of t into the equation and calculate the corresponding y-values. For example, you can substitute t = 0, 1, -1, etc., and calculate the y-values accordingly.
To find the x-intercepts, you need to solve the equation for t when y = 0. Setting 1/4t(t-2)^2 = 0 and solving for t will give you the values of t at which the graph intersects the x-axis. This will give you the x-intercepts of the graph.