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Find the vertex of each parabola

y=x^2-4x+2

y=4-6x-3x^2


y=2x^+8x-13

User Erik Rose
by
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1 Answer

5 votes

Answer:

1. y = x^2 + 6x + 2 = x^2 + 6x + 9 - 7 = (x + 3)^2 - 7

The vertex of the parabola whose equation is y = x^2 + 6x + 2 is (-3, -7).

2. y = 3x^2 + 24x - 1 = 3(x^2 + 8x) - 1 = 3(x^2 + 8x + 16) - 1 - 48 = 3(x + 4)^2 - 49

The line of symmetry for the parabola whose equation is y = 3x^2 + 24x - 1 is x = -4

3. y = x^2 + 10x + 25 = (x + 5)^2

The line of symmetry for the parabola whose equation is y = x^2 + 10x + 25 is x = -5

4. y = -2x^2 + 8x - 5 = -2(x^2 - 4x) - 5 = -2(x^2 - 4x + 4) - 5 + 8 = -2(x - 2)^2 + 3

The vertex of the parabola whose equation is y = -2x^2 + 8x - 5 is (2, 3).

5. y = x^2 - 12x + 7 = x^2 - 12x + 36 - 29 = (x - 6)^2 - 29

The line of symmetry for the parabola whose equation is y = x^2 - 12x + 7 is x = 6

6. y = x^2 - 4x + 6 = x^2 - 4x + 4 + 2 = (x - 2)^2 + 2

The vertex of the parabola whose equation is y = x^2 - 4x + 6 is (2, 2).

User Sam Chen
by
6.0k points
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