Ask Question
23 October, 03:20

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

+1
Answers (1)
  1. 23 October, 03:30
    0
    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).
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “What are the x-intercepts, rounded to the nearest tenth if necessary, of the function y = x2 - 15? ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers