Question number 10 please answer fast it is urgent
1 answer:
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The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at (-1.5, -3.25).
_____
For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
Answer:
B(-1, 0)
Step-by-step explanation:
coordinates of midpoint p(x,y) are calculated as
---------- (1)
-----------(2)
Here and
<u>Using equation (1)</u>
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<u>Using equation (2)</u>
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