What are the x-intercepts, rounded to the nearest tenth if necessary, of the function y = x2 – 15?

Hello!

To find the x-intercepts of the function, y = x² - 15, we need to make the y-value of the function equal to zero, and then we can solve for the x-intercepts.

0 = x² - 15

There are many ways to approach this question. You could factor, take the square root, use the quadratic formula, or graph the equation.

In standard form, a normal quadratic equation would be like this: y = Ax² + Bx + C. But, in this equation, there is no "Bx" term. So, the easiest way to solve for the x-intercepts is to take the square root of both sides.

1. Isolate the variable to one side of the equation.

0 = x² - 15 (add 15 to both sides)

15 = x²

2. Take the square root of both sides.

√15 = √x²

x = ±√15

√15 is about 3.87298..., which is rounded to 3.9.

-√15 is about -3.97298..., which is rounded to -3.9.

Therefore, the x-intercepts of the function, y = x² -15, are (3.9, 0) and (-3.9, 0).

12.4, 12.387, 12.26, 12.252, 6.241, 6.2, 6.015, 6.009, 1.411, 1.2, 1.056, 1.007, 0.31,   0.13, 0.5, 0.04

step-by-step explanation:

ur welcome : )

mr. india