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.
which 3 are correct?
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I see only 2 of them correct.
Try finding the vertex and evaluating b^2 -4ac
To find the vertex of the quadratic equation y = 4t^2 + 16t + 18, we can use the formula x = -b/(2a). In this case, a = 4 and b = 16. Plugging in these values, we get:
t = -16/(2*4)
t = -16/8
t = -2
To find the y-coordinate of the vertex, we substitute the value of t into the equation:
y = 4(-2)^2 + 16(-2) + 18
y = 4(4) - 32 + 18
y = 16 - 32 + 18
y = 2
Therefore, the vertex of the quadratic equation is (-2, 2). Since the vertex has a minimum, option A is correct.
To evaluate b^2 - 4ac, we substitute the values of a, b, and c from the equation into the formula:
b^2 - 4ac = (16)^2 - 4(4)(18)
b^2 - 4ac = 256 - 288
b^2 - 4ac = -32
Since the value of b^2 - 4ac is negative, the equation 4t^2 + 16t + 18 = 6 does not have any real solutions. Therefore, option C is incorrect.
Now, let's analyze the remaining options:
B. The graph of y crosses the horizontal axis at t = 10 and t = 20: To determine if the graph crosses the horizontal axis, we need to substitute these values into the equation and check if the resulting y-values are equal to 0.
y = 4(10)^2 + 16(10) + 18
y = 400 + 160 + 18
y = 578
y = 4(20)^2 + 16(20) + 18
y = 1600 + 320 + 18
y = 1938
Since the y-values at t = 10 and t = 20 are not equal to 0, the graph does not cross the horizontal axis at those points. Therefore, option B is incorrect.
D. The function has a maximum at t = 10: To determine if the function has a maximum at t = 10, we can analyze the concavity of the graph. Since the coefficient of t^2 is positive (4), the parabola opens upwards and the vertex represents the minimum point. Therefore, option D is incorrect.
E. The vertex is at (-2, 2): We already found that the vertex of the quadratic equation is (-2, 2). Therefore, option E is correct.
F. The function could represent the height y (in meters) of a particle moving under gravity, as a function of the time t (in seconds): Without any additional information about the function, we cannot determine if it represents the height of a particle moving under gravity. Therefore, option F is incorrect.
G. The graph intercepts the vertical axis at y = 18: To find the y-intercept, we substitute t = 0 into the equation:
y = 4(0)^2 + 16(0) + 18
y = 0 + 0 + 18
y = 18
Since the y-intercept is 18, option G is correct.
H. The slope of the graph of y is always negative: The slope of the graph is determined by the coefficient of t, which is 16. Since 16 is positive, the slope of the graph is positive, not negative. Therefore, option H is incorrect.
In summary, the three correct options are:
A. The graph has a minimum.
E. The vertex is at (-2, 2).
G. The graph intercepts the vertical axis at y = 18.