Find the equation of the following parabola which has vertex at (4,−1) and 𝑥-intercepts at 𝑥=12√+4 and 𝑥=4−12√.

The equation of a parabola in vertex form is given by y = a(x-h)² + k, where (h, k) is the vertex of the parabola.

Given that the vertex of the parabola is (4, -1), we can substitute h = 4 and k = -1 into the equation to get y = a(x-4)² - 1.

The x-intercepts of the parabola are given by x = 12√ + 4 and x = 4 - 12√. These are the roots of the equation, so we can set y = 0 and solve for a.

Setting x = 12√ + 4 in the equation, we get:

0 = a((12√ + 4) - 4)² - 1 0 = a(12√)² - 1 0 = 144a - 1

Setting x = 4 - 12√ in the equation, we get:

0 = a((4 - 12√) - 4)² - 1 0 = a(-12√)² - 1 0 = 144a - 1

Since both equations are equal to zero, we can set them equal to each other:

144a - 1 = 144a - 1

This equation is always true, so there are infinitely many solutions for a. Therefore, the equation of the parabola is y = a(x-4)² - 1 for any real number a.