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Completing the square when a is negative

c)Convert to standard form by completing the square: y=-2x^2 + 3x -2

Step 1: In this case, a=-2, so factor out the -2 term, but only for the x terms:

y=-2(x^2 - (3/2)x) - 2

Now take half of the b term (in this case, b=(3/2)) and square it.

(3/2/2)^2=(3/4)^2=(9/16)

Step 2: Important: Add that number to the equation inside the brackets. Note that we actually added -2*(9/16)=(-18/16)=(-9/8) to this equation, because there is a -2 term outside of the brackets. We have to now add (9/8) to the equation outside the brackets. Why? So that you are actually just adding 0 to the equation.

y=-2(x^2 - (3/2)x + 9/16) -2 +(9/8)

Step 3: You now have a perfect square (because (-3/4)+(-3/4)=(-3/2) and (-3/4)*(-3/4)=(9/16), so group those terms together and factor:

y=-2(x^2 - (3/2)x + 9/16) -2 +(9/8)
y=-2(x - (3/4))^2 - (7/8)

Or, in decimal form:
y=-2(x - 0.75)^2 - 0.875Parabola: y=-2(x-0.75)^2-0.875

This is a parabola in standard form
with a=-2, and the vertex is at (p,q)=(0.75,-0.875).