What is (x-4)^2=6(y+1/3) ?

The solution to (x-4)^2=6(y+1/3) is Parabola with (h,k)=(4,-1/3),p= 3/2

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(x-4)^2=6(y+1/3)

(x - 4)^{2} = 6 (y + \frac{1}{3})

(h, k) = (4, - \frac{1}{3}), p = \frac{3}{2}

Solution steps

(x - 4)^{2} = 6 (y + \frac{1}{3})

Rewrite

(x - 4)^{2} = 6 (y + \frac{1}{3})

in the standard form:

4 \cdot \frac{3}{2} (y - (- \frac{1}{3})) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, - \frac{1}{3}), p = \frac{3}{2}

5 f (x) = x^{2} - 4 x + 2

5 x^{2} - 50 x - 4 y + 113 = 0

x^{2} + 2 x - 5 y = 6

21 - 3 x = y^{2} + 6 y + 9

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

What is (x-4)^2=6(y+1/3) ?
The solution to (x-4)^2=6(y+1/3) is Parabola with (h,k)=(4,-1/3),p= 3/2