What is (1-p(x))/3-(7x)/6+x^2=0 ?

The solution to (1-p(x))/3-(7x)/6+x^2=0 is Parabola with (h,k)=(7/12 ,-49/48),p= 1/12

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(1-p(x))/3-(7x)/6+x^2=0

\frac{1 - p ( x )}{3} - \frac{7 x}{6} + x^{2} = 0

(h, k) = (\frac{7}{12}, - \frac{49}{48}), p = \frac{1}{12}

Solution steps

\frac{1 - y}{3} - \frac{7 x}{6} + x^{2} = 0

Rewrite

\frac{1 - y}{3} - \frac{7 x}{6} + x^{2} = 0

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{49}{48})) = (x - \frac{7}{12})^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{12}, - \frac{49}{48}), p = \frac{1}{12}

x - y^{2} = - 9

3 x^{2} + 4 x + y + 2 = 0

(X + 10)^{2} = 3 (y - 12)

x - y^{2} = - 3

x^{2} = \frac{6 r ( x )}{3} - \frac{1}{2}

What is (1-p(x))/3-(7x)/6+x^2=0 ?
The solution to (1-p(x))/3-(7x)/6+x^2=0 is Parabola with (h,k)=(7/12 ,-49/48),p= 1/12