Question

1. find the x-intercepts of $y = x^2 + 25$.

\bigcirc $x = \pm 25$
\bigcirc there is no real number solution, so there is no x-intercept.
\bigcirc $x = \pm 5$
\bigcirc $x = \pm 25i$

Explanation:

Step 1: Recall the definition of x - intercepts

To find the x - intercepts of a function \(y = f(x)\), we set \(y = 0\) and solve for \(x\). So for the function \(y=x^{2}+25\), we set \(y = 0\), which gives the equation \(x^{2}+25=0\).

Step 2: Solve the equation for x

We can rewrite the equation as \(x^{2}=- 25\). In the set of real numbers, the square of any real number \(x\) (i.e., \(x^{2}\)) is always non - negative (because if \(x\) is real, \(x^{2}\geq0\) for all real \(x\)). But \(-25\) is negative. So there are no real numbers \(x\) that satisfy the equation \(x^{2}=-25\). Since x - intercepts are found by finding the real values of \(x\) when \(y = 0\), and there are no real solutions for \(x\) in the equation \(x^{2}+25 = 0\), there are no x - intercepts.

Answer:

There is no real number solution, so there is no x - intercept.