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# How do I use the vertex formula to determine the vertex of the graph for ## x^2 + 6x + 8y + 25 = 0##?

In vertex form the equation looks like

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

So the vertex is the point ##(-3,-2)##

This has always been kinda confusing for me since no one explained to me so I learned on my own. I found this graphing calculator online which shows the vertex,explanations & everything! I will put the link below....

http://www.mathportal.org/calculators/quadratic-equation/quadratic-function-grapher.php

Algebraically, to put the equation in vertex form ##y-y_1=(x-x_1)^2##

we need to solve the equation for ##y## in terms of ##x##.

##x^2+6x+8y+25=0## ##8y=-x^2-6x-25## ##y=-1/8(x^2+6x)-25/8##

Completing the square inside the parentheses gives us

##y=-1/8(x^2+6x+9)-25/8+9/8## ##y=-1/8(x+3)^2-2##

or

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

Converting to vertex form we get

##y-(-2)=-1/8(x-(-3))^2## which gives us the the vertex ##(-3,-2)##.

The graph is below.