Find the x-intercepts of the parabola with vertex (1, -9) and y intercept at (0, -6).

A.
(-1, 0), (3, 0)

B.
(-0.73, 0), (2.73, 0)

C.
(-1.48, 0), (2.48, 0)

D.
(-4.67, 0), (1.67, 0)

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GoddessofDork GoddessofDork · Answer 1 · 2018-09-06T02:25:45+02:00

So firstly, plug the vertex into the vertex form: [tex] y=a(x-1)^2-9 [/tex]

Next, we first need to solve for a. Plug (0,-6) into the equation to solve for a as such:

[tex] -6=a(0-1)^2-9\\-6=a(-1)^2-9\\-6=a-9\\3=a [/tex]

Now we know that our equation is y = 3(x - 1)^2 - 9.

Now to solve for the zeros (x-intercepts). Set y to 0 and add 9 to both sides of the equation: [tex] 9=3(x-1)^2 [/tex]

Next, divide both sides by 3: [tex] 3=(x-1)^2 [/tex]

Next, square root both sides: [tex] \pm\sqrt{3}=x-1 [/tex]

Next, add 1 to both sides: [tex] 1\pm\sqrt{3}=x\\2.73,-0.73=x [/tex]

Your x-intercepts are (-0.73,0) and (2.73,0), or B.

jainveenamrata jainveenamrata · Answer 2 · 2022-01-28T05:58:58+01:00

The x-intercept is [tex]1 \pm \sqrt{3}[/tex].

When the power of variable x is 2.

Given

Vertex on (1, -9)

point (0, -6)

We know the equation of a parabola is [tex]y= a(x-h)^{2} +k[/tex]

As (h. k) is (1, -9) then the equation becomes [tex]y= a(x-1)^{2} -9[/tex]

Parabola passing through the point (0, -6)

[tex]\begin{aligned} y &= a(x-1)^{2} -9\\-6 &= a(0 -1)^{2} -9\\a &= 3\\\end{aligned}[/tex]

Then equation becomes [tex]y= 3(x-1)^{2} -9[/tex]

For x-intercept, the point becomes (t, 0). then

[tex]y &= 3(x-1)^{2} -9\\0 &= 3(t-1)^{2} -9\\9 &= 3(t-1)^{2} \\3 &= (t-1)x^{2} \\t-1 &= \sqrt{3} \\t &= 1 \mp \sqrt{3}[/tex]

Hence, the x-intercept is [tex]1 \pm \sqrt{3}[/tex].

More about the parabola link is given below.

https://brainly.com/question/8495268