What is 4x^2-12x+q(x)=0 ?

The solution to 4x^2-12x+q(x)=0 is Parabola with (h,k)=(3/2 ,9),p=-1/16

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4x^2-12x+q(x)=0

4 x^{2} - 12 x + q (x) = 0

(h, k) = (\frac{3}{2}, 9), p = - \frac{1}{16}

Solution steps

4 x^{2} - 12 x + y = 0

Rewrite

4 x^{2} - 12 x + y = 0

in the standard form

4 (- \frac{1}{16}) (y - 9) = (x - \frac{3}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{2}, 9), p = - \frac{1}{16}

x = y^{2} - 9 x + 3

(1 - x)^{2} + (2 - y)^{2} - (7 - x)^{2} + y^{2} = 40

(x + 7)^{2} = 6 (y - 4)

(y - 6)^{2} = - 12 (x + 4)

9 x^{2} + 18 x - 3 = 36 r (x)

What is 4x^2-12x+q(x)=0 ?
The solution to 4x^2-12x+q(x)=0 is Parabola with (h,k)=(3/2 ,9),p=-1/16