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y= 4t^2 + 16t + 18

A. The graph has a minimum.
B. The graph of y crosses the horizontal axis at t = 10 and t = 20.
C. The equation 4t^2 + 16 + 18 = 6 has one real solution.
D. The function has a maximum at t = 10
E. The vertex is at (-2, 2)
F. The function could represent the height y (in metres) of a particle moving under gravity, as a function of the time t (in seconds).

G. The graph intercepts the vertical axis at y = 18
H. The slope of the graph of y is always negative.

I think the answers are G, E and A. Is this correct

consider 4t^2 + 16t + 18=6

4t^2 + 16t + 18-6=0
b^2-4ac = 256-16*12 is a positive, so there are two real solutions.

so c is wrong.

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