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Answer:
(7, 0) and (-5, 0)
Step-by-step explanation:
Vertex form
[tex]y=a(x-h)^2+k[/tex]
(where (h, k) is the vertex)
Given:
[tex]\implies y=a(x-1)^2-108[/tex]
Given:
[tex]\implies a(0-1)^2-108=-105[/tex]
[tex]\implies a(-1)^2=-105+108[/tex]
[tex]\implies a=3[/tex]
Therefore:
[tex]\implies y=3(x-1)^2-108[/tex]
The x-intercepts are when y = 0
[tex]\implies 3(x-1)^2-108=0[/tex]
[tex]\implies 3(x-1)^2=108[/tex]
[tex]\implies (x-1)^2=36[/tex]
[tex]\implies x-1=\pm \sqrt{36}[/tex]
[tex]\implies x=1\pm 6[/tex]
[tex]\implies x=7, x=-5[/tex]
Therefore, the x-intercepts are (7, 0) and (-5, 0)