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).