I am trying to find the y-intercept

The equation is x+1 = -2(y+5)^2

This is a parabola, but not exactly a function (does not pass vertical line test)

How exactly do I find the y-intercept? If I put x to 0 like the traditional way, than there is no solution. But when I graph this out, the y-intercept exists. How is this so? Is there a proper way to find the y-intercept? Have I made any errors?

Thanks

Guest Aug 1, 2019

#1**0 **

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!)

ElectricPavlov Aug 1, 2019