What is (2x^2-28x)/(-k(x)-78)=1 ?

The solution to (2x^2-28x)/(-k(x)-78)=1 is Parabola with (h,k)=(7,20),p=-1/8

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(2x^2-28x)/(-k(x)-78)=1

\frac{2 x ^{2} - 28 x}{- k ( x ) - 78} = 1

(h, k) = (7, 20), p = - \frac{1}{8}

Solution steps

\frac{2 x ^{2} - 28 x}{- y - 78} = 1

Rewrite

\frac{2 x ^{2} - 28 x}{- y - 78} - 1 = 0

in the standard form

4 (- \frac{1}{8}) (y - 20) = (x - 7)^{2}

Therefore parabola properties are:

(h, k) = (7, 20), p = - \frac{1}{8}

y = - \frac{1}{12} x^{2} - \frac{4}{3} x - \frac{5}{2}

x = 2 \cdot y (y - 2)

h (x) = x - 0.5 x^{2}

y^{2} + z = 2

\frac{x}{y} = y

What is (2x^2-28x)/(-k(x)-78)=1 ?
The solution to (2x^2-28x)/(-k(x)-78)=1 is Parabola with (h,k)=(7,20),p=-1/8