Groph the quadratic \( y=x^{2}+3 x+8 \). What is the \( x \)-intercept(s)? (1 point)

To find the \( x \)-intercepts of the quadratic \( y = x^2 + 3x + 8 \), set \( y = 0 \): \[ x^2 + 3x + 8 = 0 \] Using the quadratic formula, where \( a = 1, b = 3, c = 8 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Calculating the discriminant: \[ b^2 - 4ac = 3^2 - 4(1)(8) = 9 - 32 = -23 \] Since the discriminant is negative, this quadratic does not cross the x-axis, meaning there are no \( x \)-intercepts. The graph opens upward (since the coefficient of \( x^2 \) is positive), and the vertex will be the lowest point on the graph, but it will never touch or cross the x-axis.