x+1 =-2(y^2+10y+25)
x = -2y^2-20y-51 x will equal 0 at the y intercepts
0 = -2y^2-20y-51 now use the Quadratic Formula to find the y's
\(y = {-b \pm \sqrt{b^2-4ac} \over 2a}\) a = -2 b = -20 c = -51 Let me know what you find !
(Edit: 'cause when I graph it, I find NO y-axis intercepts!)