Expert: Sherry Wallin - 4/17/2011

Question
Write the standard equation of the following:

y=-4x^2+3x-2

Answer
This seems to be in the standard form of a quadratic equation. The standard form of a 2nd degree (quadratic) equation: is ax^2+bx+c = 0 but perhaps they want it in y = a(x-h)^2 + k. Then you need to complete the square:
But in order to complete the square you must have a one as the coefficient on x^2 so factor out a -4:
y = -4[x^2-(3/4)x +(1/2)] and now complete the square on the part in the square brackets:
y = -4(x^2-(3/4)x        ) -4(1/2)
y = -4(x^2-(3/4)x +(-3/8)^2) -4(1/2) -[-4(-3/8)^2
y = -4(x^2-(3/4)x +(-3/8)^2) -2 +4(9/64)
y = -4(x^2-(3/4)x +(-3/8)^2) -2 +9/16
y = -4(x-(3/8))^2 -32//16+9/16
y = -4(x-(3/8))^2 -23/16
so your h is 3/8 and your k is -23/16 which is your vertex of the parabola


Math Prof

Notice if you just write -4x^2+3x-2 = 0 and use x = -b/2a you get -3/(2(-4)) = 3/8
and f(3/8) is -4(-3/8)^2 +3(3/8) -2
=  -4(9/64) +9/8 -2
= -9/16 +18/16 - 32/16
= -23/16
and the vertex is (-b/2a, f(-b/2a)) => (3/8, -23/16)