What is (8(8+1))/2 >= 8*q+p-(q(q-1))/2 ?

The solution to (8(8+1))/2 >= 8*q+p-(q(q-1))/2 is Parabola with (h,k)=(17/2 ,-1/8),p= 1/2

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(8(8+1))/2 >= 8*q+p-(q(q-1))/2

\frac{8 ( 8 + 1 )}{2} \geq 8 \cdot q + p - \frac{q ( q - 1 )}{2}

(h, k) = (\frac{17}{2}, - \frac{1}{8}), p = \frac{1}{2}

Solution steps

\frac{8 ( 8 + 1 )}{2} \geq 8 q + p - \frac{q ( q - 1 )}{2}

Rewrite

\frac{8 ( 8 + 1 )}{2} \geq 8 q + p - \frac{q ( q - 1 )}{2}

in the standard form

4 \cdot \frac{1}{2} (p - (- \frac{1}{8})) = (q - \frac{17}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{17}{2}, - \frac{1}{8}), p = \frac{1}{2}

y > - 2 (x - 3)^{2} + 4

f + 4 g = \frac{1}{5} d 51 d

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

p (6) = - 11 s^{2} - 8

x^{2} + y = - 4 x + 1

What is (8(8+1))/2 >= 8*q+p-(q(q-1))/2 ?
The solution to (8(8+1))/2 >= 8*q+p-(q(q-1))/2 is Parabola with (h,k)=(17/2 ,-1/8),p= 1/2