What is 1/7 (y-15)=(x+3)^2 ?

The solution to 1/7 (y-15)=(x+3)^2 is Parabola with (h,k)=(-3,15),p= 1/28

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1/7 (y-15)=(x+3)^2

\frac{1}{7} (y - 15) = (x + 3)^{2}

(h, k) = (- 3, 15), p = \frac{1}{28}

Solution steps

\frac{1}{7} (y - 15) = (x + 3)^{2}

Rewrite

\frac{1}{7} (y - 15) = (x + 3)^{2}

in the standard form:

4 \cdot \frac{1}{28} (y - 15) = (x - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 3, 15), p = \frac{1}{28}

yy - (- 4) = 3 (x - (- 2))

21 x - 32 y^{2} = 4 - 7 y

x^{2} - 14 x + 4 y + 45 = 0

(x + 4)^{2} = 12 (y - 1)

2 x^{2} + 3 y = 7

What is 1/7 (y-15)=(x+3)^2 ?
The solution to 1/7 (y-15)=(x+3)^2 is Parabola with (h,k)=(-3,15),p= 1/28