Question

How do you find the vertex and intercepts for `y^2 + 6y + 8x + 25 = 0`?

Answer 1

As explained below.

Explanation:

Write the equation as `8x=-(y^2 +6y) -25`

Or, `8x= - (y^2 +6y +9-9) -25`

`8x= -(y+3)^2 -16`

`x= -1/8 (y+3)^2 -2`

Vertex is(-2, -3)

x Intercept would be when y=0. Then x=`-9/8-2=-25/2`

y intercept would be when x=0, then quadratic equation `y^2 +6y+25=0` would yield imaginary values for y. Which means that the horizontal parabola would not cross y axis, hence no y intercept.
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