To find the x-intercepts of the quadratic equation \(y = x^2 + 3x + 8\), we need to set \(y\) to 0 and solve for \(x\):
\[ 0 = x^2 + 3x + 8 \]
To find the x-intercepts, we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \(x^2 + 3x + 8\):
- \(a = 1\)
- \(b = 3\)
- \(c = 8\)
Now substituting these values into the quadratic formula:
- Calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 3^2 - 4(1)(8) = 9 - 32 = -23 \]
Since the discriminant is negative (\(-23\)), there are no real solutions for \(x\). This means that the quadratic does not cross the x-axis, and therefore there are no x-intercepts.
Thus, the answer is:
There is no x-intercept.