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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>e. the graph has a minimu...Question
in the xy-coordinate plane, the graph of the equation y=2x^2-12x-32 has zeros x=d and x=e, where d>e. the graph minimum at (f, -50). what are the values of d,e, and f?
Answers
GPT 3.5
To find the values of d, e, and f, we first need to find the x-values of the zeros and the x-value of the minimum point.
1. Zeros:
To find the zeros of the equation y=2x^2-12x-32, we set y=0:
0 = 2x^2-12x-32
Divide by 2 to simplify:
0 = x^2 - 6x - 16
Factor the quadratic equation:
0 = (x-8)(x+2)
This gives us x=8 and x=-2.
Since it is given that d>e, d=8 and e=-2.
2. Minimum point:
To find the x-value of the minimum point, we use the formula for the x-coordinate of the minimum of a parabola:
f = -(-b) / (2a)
f = 12 / (2*2)
f = 6
Therefore, the values are:
d = 8
e = -2
f = 6
1. Zeros:
To find the zeros of the equation y=2x^2-12x-32, we set y=0:
0 = 2x^2-12x-32
Divide by 2 to simplify:
0 = x^2 - 6x - 16
Factor the quadratic equation:
0 = (x-8)(x+2)
This gives us x=8 and x=-2.
Since it is given that d>e, d=8 and e=-2.
2. Minimum point:
To find the x-value of the minimum point, we use the formula for the x-coordinate of the minimum of a parabola:
f = -(-b) / (2a)
f = 12 / (2*2)
f = 6
Therefore, the values are:
d = 8
e = -2
f = 6