17-Identify the vertex, axis of symmetry and x-intercept(s) of F(x)=(x+5)^2-6. I have no idea how the x intercepts are -5 +/- sq. root 6 and 0. can you show me how this came to be? I factored it so it would be x^2+10x+19, but there's no common factors in ten and 19. Can you show me, please?

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17-Identify the vertex, axis of symmetry and x-intercept(s) of F(x)=(x+5)^2-6. I have no idea how the x intercepts are -5 +/- sq. root 6 and 0. can you show me how this came to be? I factored it so it would be x^2+10x+19, but there's no common factors in ten and 19. Can you show me, please?

Answer 1 · 2009-11-08 04:07:00

for the x-intercepts, F(x) = 0. (the y = 0)

so (x+5)^2-6 = 0
(x+5)^2 = 6
x+5 = ± √6
x = -5 ± √6

You expanded the function, you did not factor it.
Once you get x^2 + 10x + 19 = 0,
if you use the formula and simplify the answer, it comes out the same as my answer.
I have no idea where you got the 0, a parabola can have at most 2 x-intercepts.

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17-Identify the vertex, axis of symmetry and x-intercept(s) of F(x)=(x+5)^2-6. I have no idea how the x intercepts are -5 +/- sq. root 6 and 0. can you show me how this came to be? I factored it so it would be x^2+10x+19, but there's no common factors in ten and 19. Can you show me, please?

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