What is 5y+4=((x+3)^2+1)/2 ?

The solution to 5y+4=((x+3)^2+1)/2 is Parabola with (h,k)=(-3,-7/10),p= 5/2

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5y+4=((x+3)^2+1)/2

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

(h, k) = (- 3, - \frac{7}{10}), p = \frac{5}{2}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{5}{2} (y - (- \frac{7}{10})) = (x - (- 3))^{2}

Therefore parabola properties are:

(h, k) = (- 3, - \frac{7}{10}), p = \frac{5}{2}

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

6 y^{2} - 18 x = 0

x (x + 1) - 3 = - y

x^{2} - 8 x + 4 y - 20 = 0

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

What is 5y+4=((x+3)^2+1)/2 ?
The solution to 5y+4=((x+3)^2+1)/2 is Parabola with (h,k)=(-3,-7/10),p= 5/2