What is 2x^2+5x-k(x)=0 ?

The solution to 2x^2+5x-k(x)=0 is Parabola with (h,k)=(-5/4 ,-25/8),p= 1/8

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2x^2+5x-k(x)=0

2 x^{2} + 5 x - k (x) = 0

(h, k) = (- \frac{5}{4}, - \frac{25}{8}), p = \frac{1}{8}

Solution steps

2 x^{2} + 5 x - y = 0

Rewrite

2 x^{2} + 5 x - y = 0

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{25}{8})) = (x - (- \frac{5}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{5}{4}, - \frac{25}{8}), p = \frac{1}{8}

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

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

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}

What is 2x^2+5x-k(x)=0 ?
The solution to 2x^2+5x-k(x)=0 is Parabola with (h,k)=(-5/4 ,-25/8),p= 1/8