Question
In the xy-coordinate plane, the graph of the equation y=2x^2-12x-32 has zeros at x=d and x=e, where d is greater than e. The graph has a minimum at (f,-50). What are the values of d, e, and f?
Answers
GPT 3.5
To find the zeros of the equation y = 2x^2 - 12x - 32, we need to set y equal to zero and solve for x:
0 = 2x^2 - 12x - 32
We can factor this quadratic equation:
0 = 2(x^2 - 6x - 16)
0 = 2(x - 8)(x + 2)
Setting each factor equal to zero, we find the zeros:
x = 8 and x = -2
Since d is greater than e, d = 8 and e = -2.
To find the x-coordinate of the minimum, we can use the formula x = -b/2a, where a = 2 (the coefficient of x^2) and b = -12 (the coefficient of x):
f = -(-12) / (2*2) = 12 / 4 = 3
Therefore, d = 8, e = -2, and f = 3.
0 = 2x^2 - 12x - 32
We can factor this quadratic equation:
0 = 2(x^2 - 6x - 16)
0 = 2(x - 8)(x + 2)
Setting each factor equal to zero, we find the zeros:
x = 8 and x = -2
Since d is greater than e, d = 8 and e = -2.
To find the x-coordinate of the minimum, we can use the formula x = -b/2a, where a = 2 (the coefficient of x^2) and b = -12 (the coefficient of x):
f = -(-12) / (2*2) = 12 / 4 = 3
Therefore, d = 8, e = -2, and f = 3.