Find the x intercepts of the parabola with vertex (2,13) and y-intercept (0,5) write your answer in this form: (x1,,y2) if necessary, round to the nearest hundredth

The standard formula for a parabola are

(x-h)^2 = +/- 4a (y-k) or (y-k)^2 = +/- 4a(x-h)

where
(h,k) is the coordinates of the vertex
a is the length of the focus from the vertes
+4a if the parabola opens upwards or to the right
-4a if the parabola opens downwards or to the left

The vertex (2,13) is situated in the 1st quadrant of the Cartesian plane. It only has y-intercept. This means that it only passes the y-axis once. Therefore, the parabola must open downwards and it passes the x-axis twice. The intersections at the x-axis are the x-intercepts. If the parabola has 2 x-intercepts, then the equation would be (x-h)^2 = -4a(y-k).

Let's use the y-intercept (0,5) to determine 4a:

(0-2)^2=-4a(5-13)
4a = 0.5

Therefore, the equation of the parabola is (x-2)^2 = -1/2(y-13). To find the x-intercepts, let y=0.

(x-2)^2 = -1/2(0-13) = 6.5
x-2 = +/- √6.5 = +/- 2.55
x = 4.55 and and -0.55

The x-intercepts are (-0.55,0) and (4.55,0).

C. 40,320 just did the math right now