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October 24, 2014

October 24, 2014

Posted by **Cassie** on Saturday, June 6, 2009 at 2:19pm.

a) vertex (2,3); x = -2

b) vertex (-2,-3); x = 2

c) vertex (2,3); x = 2

d) vertex (-2,-3); x = -2

I have no idea how to solve this problem! Please help. Thank you! :)

- Algebra -
**MathMate**, Saturday, June 6, 2009 at 3:51pmAll parabolas of the form

y=ax^{2}+ bx + c

have their axis of symmetry passing through the vertex.

Thus a and b can be eliminated on that basis.

Note that when the x^{2}term is positive, the parabola has a minimum at the vertex.

By inspection, when x=2, the value of y is at its minimum because the 5(x-2)^{2}term is zero. Any other value of x will render the term > 0. Thus the vertex is at x=2, where y=3.

Thus the answer is (C).

Check:

c. y=5(2-2)^{2}+3=3

d. y=5(-2-2)^{2}+3=83 (not equal to 3)

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