Answer:
[tex]x=1+\sqrt{3}[/tex] and [tex]x=1-\sqrt{3}[/tex]
Step-by-step explanation:
We have been given the parabola with vertex (1, -9) and y intercept at (0, -6).
Now we need to find the x-intercepts of that parabola. So first we begin by finding the equation of parabola using vertex formula:
[tex]y=a\left(x-h\right)^2+k[/tex]
Vertex for this formula is given by (h,k)
Compare that with given vertex (1,-9), we get: h=1, k=-9
So plug these into vertex formula:
[tex]y=a\left(x-1\right)^2-9[/tex]...(i)
Plug given point (0, -6). into (i)
[tex]-6=a\left(0-1\right)^2-9[/tex]
[tex]-6=a\left(-1\right)^2-9[/tex]
[tex]-6=a\left(1\right)-9[/tex]
[tex]-6+9=a[/tex]
[tex]3=a[/tex]
Plug a=3 into (i)
[tex]y=3\left(x-1\right)^2-9[/tex]
Now to find x-intercept, we just plug y=0 and solve for x
[tex]y=3\left(x-1\right)^2-9[/tex]
[tex]0=3\left(x-1\right)^2-9[/tex]
[tex]9=3\left(x-1\right)^2[/tex]
[tex]3=\left(x-1\right)^2[/tex]
[tex]\pm \sqrt{3}=x-1[/tex]
[tex]1+\pm \sqrt{3}=x[/tex]
Hence final answer are
[tex]x=1+\sqrt{3}[/tex] and [tex]x=1-\sqrt{3}[/tex]
