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Q: Find the vertex of y equals x2 plus 10x plus 22?

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(5,-5)

y=x2-10x+30=(x-5)2-25+30=(x-5)2+5

You find x. Oh! There it is!! ^ See?

You only need to find the vertex of a parabola, so I assume this is the equation. Y = X^2 - 10X + 25 set to 0 X^2 - 10X + 25 = 0 now, you need to complete the square. Halve the linear term (-10), square it and add that to both sides. subtract 25 from each side to start X^2 - 10X = - 25 X^2 - 10X + 25 = - 25 + 25 factor left; gather like terms right (X - 5)^2 = 0 Vertex (5,0)

X2 + 10X - 13 = 17 add 13 to each side X2 + 10X = 30 halve the linear coefficient ( 10 ), square it and add it to both sides X2 + 10X + 25 = 30 + 25 factor on the left and gather terms on the right (X + 5)2 = 55 (X + 5)2 - 55 = 0 ---------------------------vertex form, (- 5, - 55 ), is the vertex -------------------------------------------[-------------------------------]

x2 + 10x + 21 = (x + 3)(x + 7)

x + 13x + 10x = 50 - 6

10x + 3 = 15 10x = 15 - 3 10x = 12 x = 1.2

y=-10x-4

5x + 4 = 10x - 5 10x - 5x = 4 + 5 5x = 9 x = 1.8

x=-1.6

6.28

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