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?
To find the values of d, e, and f, we first need to find the x-coordinates of the zeros and the x-coordinate of the minimum point.
1. Finding the zeros:
To find the zeros of the equation y = 2x^2 - 12x - 32, we need to set y = 0 and solve for x.
0 = 2x^2 - 12x - 32
2(x^2 - 6x - 16) = 0
2(x - 8)(x + 2) = 0
Setting each factor to zero gives us:
x - 8 = 0 or x + 2 = 0
x = 8 or x = -2
So, the zeros are at x = 8 and x = -2.
2. Finding the x-coordinate of the minimum point:
The x-coordinate of the minimum point can be found using the formula x = -b/(2a), where the equation is in the form y = ax^2 + bx + c.
In this case, a = 2, b = -12, and c = -32.
x = -(-12)/(2*2)
x = 12/4
x = 3
So, the x-coordinate of the minimum point is 3.
Since d is greater than e, the values of d, e, and f are:
d = 8
e = -2
f = 3
Therefore, the values of d, e, and f are 8, -2, and 3 respectively.