Consider the graph of the following quadratic equation. y=–x^2–10x+24 what is the y value of the vertex?

In standard form, y=Ax^2+Bx+C, you can't read the vertex easily, so you can complete the square to get vertex form, y=a(x-h)^2 +k where (h,k) is the vertex...

This equation can be changed to
x^2-10x+25-25+24
(x-5)^2-1

Therefore, the vertex is (5,-1)

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