What is x^2+x+h(x)-(15h(x))/4 =0 ?

The solution to x^2+x+h(x)-(15h(x))/4 =0 is Parabola with (h,k)=(-1/2 ,-1/11),p= 11/16

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x^2+x+h(x)-(15h(x))/4 =0

x^{2} + x + h (x) - \frac{15 h ( x )}{4} = 0

(h, k) = (- \frac{1}{2}, - \frac{1}{11}), p = \frac{11}{16}

Solution steps

x^{2} + x + y - \frac{15 y}{4} = 0

Rewrite

x^{2} + x + y - \frac{15 y}{4} = 0

in the standard form

4 \cdot \frac{11}{16} (y - (- \frac{1}{11})) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, - \frac{1}{11}), p = \frac{11}{16}

y 5 y - 2 x = 10

\frac{- k ( x )}{3 x ^{2}} = \frac{- 6}{5}

(y - 1)^{2} = 8 (x + 2)

- 6 y^{2} + 4 y = x + 3 y - 2

5 y^{2} - z - 6 = 0

What is x^2+x+h(x)-(15h(x))/4 =0 ?
The solution to x^2+x+h(x)-(15h(x))/4 =0 is Parabola with (h,k)=(-1/2 ,-1/11),p= 11/16