Question

Which could be the graph of f of x equals x squared plus 6 x plus 11?

Answer options with 5 options
A.
On a coordinate plane, a downward opening parabola is in quadrant 2 with two x intercepts.

D.
On a coordinate plane, a downward opening parabola is in quadrant 3 with no x-intercepts.

B.
On a coordinate plane, an upward opening parabola is in quadrant 1 with no x-intercepts.

E.
On a coordinate plane, a downward opening parabola is in quadrant 4 with no x-intercepts.

C.
On a coordinate plane, an upward opening parabola is in quadrant 2 with no x-intercepts.

Answer 1

To determine the properties of the function \( f(x) = x^2 + 6x + 11 \), we can analyze it using the standard form of a quadratic function, \( ax^2 + bx + c \).

1. **Determine the orientation of the parabola:**
- The coefficient of \( x^2 \) is positive (1), which means the parabola opens upward.

2. **Check for x-intercepts:**
- To find the x-intercepts, we set \( f(x) = 0 \):
\[
x^2 + 6x + 11 = 0
\]
- We can use the discriminant to determine the nature of the roots:
\[
D = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 11 = 36 - 44 = -8
\]
- Since the discriminant is negative (\( D < 0 \)), there are no real roots. This implies that there are no x-intercepts.

3. **Determine the vertex of the parabola:**
- The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \):
\[
x = -\frac{6}{2 \cdot 1} = -3
\]
- To find the y-coordinate of the vertex, we substitute \( x = -3 \) back into the function:
\[
f(-3) = (-3)^2 + 6(-3) + 11 = 9 - 18 + 11 = 2
\]
- Therefore, the vertex is at the point (-3, 2), which is in Quadrant II.

Based on the analysis:
- The parabola opens upward (since \( a > 0 \)),
- There are no x-intercepts (since the discriminant is negative),
- The vertex is located in Quadrant II.

With these considerations, the correct match among the options provided is:

**C. On a coordinate plane, an upward opening parabola is in quadrant 2 with no x-intercepts.**