# maths

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.

Are the three options G , C and E is this correct ??

It does have a minimum, since the first derivative is zero at a location where the second derivative is positive. That location is t= -2, y = 2.

You probably stated option C incorrectly. 4t^2 +16 t +12 = 0 can be rewritten 4(t^2 + 4t +3) = 0, and that has two solutions.

It cannot represent the height of a particle moving under gravity, since the coefficient if the t^2 term is not
-g/2 = -4.9 m/s^2

G and E are two of the correct answers. C is not.

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