Help Please! Find the value of x for which ABCD must be a parallelogram.
1 answer:
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Answer:
D. Minimum at (3, 7)
Step-by-step explanation:
We can add and subtract the square of half the x-coefficient:
y = x^2 -6x +(-6/2)^2 +16 -(-6/2)^2
y = (x -3)^2 +7 . . . . . simplify to vertex form
Comparing this to the vertex for for vertex (h, k) ...
y = (x -h)^2 +k
We find the vertex to be ...
(3, 7) . . . . vertex
The coefficient of x^2 is positive (+1), so the parabola opens upward and the vertex is a minimum.
