Question

Find the x intercept(s) y= -x^2 + 2

Answer 1

Answer:

$(\pm \sqrt{2}, 0)$

Step-by-step explanation:

The x intercepts are basically the points at which y = 0.

-x^2 + 2 = 0

-x^2 = -2

x^2 = 2

x = $\pm \sqrt{2}$

Answer 2

For this case we have by definition, that to find the x-intercept points, we must make the variable y = 0 and clear the value of "x". So:

$y = -x ^ 2 + 2\\0 = -x ^ 2 + 2\\x ^ 2 = 2\\x = \sqrt {2}$

So, the x-intercepts are:

$(x_ {1}, y_ {1}) = (\sqrt {2}, 0)\\(x_ {2}, y_ {2}) = (- \sqrt {2}, 0)$

ANswer:

$(x_ {1}, y_ {1}) = (\sqrt {2}, 0)\\(x_ {2}, y_ {2}) = (- \sqrt {2}, 0)$