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

Mar 21, 2017

As explained below.

#### Explanation:

Write the equation as $8 x = - \left({y}^{2} + 6 y\right) - 25$

Or, $8 x = - \left({y}^{2} + 6 y + 9 - 9\right) - 25$

$8 x = - {\left(y + 3\right)}^{2} - 16$

$x = - \frac{1}{8} {\left(y + 3\right)}^{2} - 2$

Vertex is(-2, -3)

x Intercept would be when y=0. Then x=$- \frac{9}{8} - 2 = - \frac{25}{2}$

y intercept would be when x=0, then quadratic equation ${y}^{2} + 6 y + 25 = 0$ 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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