Which three options are true about the quadratic equation??
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.
The three correct options are G, E, and F.
G. The graph intercepts the vertical axis at y = 18. This is true because when t = 0, the equation becomes y = 4(0)^2 + 16(0) + 18 = 18.
E. The vertex is at (-2, 2). This is true because the quadratic equation is in the form y = ax^2 + bx + c, and the x-coordinate of the vertex is given by -b/2a. By substituting the values a = 4 and b = 16 into the formula, we find the x-coordinate of the vertex to be -16/(2*4) = -2. When t = -2, the equation becomes y = 4(-2)^2 + 16(-2) + 18 = 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). This is true because the given function represents a quadratic equation, and the height of a particle under gravity can be modeled by a quadratic equation.
Therefore, the correct options are G, E, and F.