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Mathematics 24 Online
OpenStudy (anonymous):

vertex of the graph of the following equation y=(x+5)^2 -2

OpenStudy (anonymous):

(-5,-2)

OpenStudy (jdoe0001):

for a parabola equation like \(\bf ax^2+bx+c\) you can find the vertex at \(\bf \left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)\)

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