Question
Groph the quadratic \( y=x^{2}+3 x+8 \). What is the \( x \)-intercept(s)? (1 point)
Ask by Page Murray. in the United States
Jan 26,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
There are no real x-intercepts for the quadratic \( y = x^{2} + 3x + 8 \).
Solution
The quadratic equation \( y = x^{2} + 3x + 8 \) can be analyzed for \( x \)-intercepts by setting \( y = 0 \):
\[
x^{2} + 3x + 8 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 3 \), and \( c = 8 \):
\[
x = \frac{-3 \pm \sqrt{9 - 32}}{2} = \frac{-3 \pm \sqrt{-23}}{2}
\]
Since the discriminant (\( \sqrt{-23} \)) is negative, there are no real \( x \)-intercepts for this quadratic function.
**Answer:**
There are no real x-intercepts because the quadratic has no real roots.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
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.
