Question

find the x-intercept(s) and vertex for the quadratic function.
f(x)= 3x^2 + 11x+5
Round to the nearest hundredth if necessary.
If there is more than one x-intercept, separate them with commas.
If applicable, click on "None"

Answer 1

To find the x-intercepts and the vertex of the quadratic function \( f(x) = 3x^2 + 11x + 5 \), we can follow these steps:

### Finding X-Intercepts

The x-intercepts occur where \( f(x) = 0 \):

\[
3x^2 + 11x + 5 = 0
\]

We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 3 \), \( b = 11 \), and \( c = 5 \).

1. Calculate the discriminant (\( b^2 - 4ac \)):

\[
b^2 - 4ac = 11^2 - 4 \cdot 3 \cdot 5 = 121 - 60 = 61
\]

2. Apply the quadratic formula:

\[
x = \frac{-11 \pm \sqrt{61}}{2 \cdot 3}
\]

This results in:

\[
x = \frac{-11 + \sqrt{61}}{6} \quad \text{and} \quad x = \frac{-11 - \sqrt{61}}{6}
\]

3. Calculate both x-intercepts:

Using a calculator:

\[
x = \frac{-11 + \sqrt{61}}{6} \approx -1.55
\]

\[
x = \frac{-11 - \sqrt{61}}{6} \approx -3.45
\]

Thus, the x-intercepts are approximately:

\[
-1.55, -3.45
\]

### Finding the Vertex

The x-coordinate of the vertex for a quadratic \( ax^2 + bx + c \) is given by:

\[
x = -\frac{b}{2a}
\]

Using the values for \( a \) and \( b \):

\[
x = -\frac{11}{2 \cdot 3} = -\frac{11}{6} \approx -1.83
\]

Now, we calculate the y-coordinate of the vertex by plugging this value back into the function:

\[
f\left(-\frac{11}{6}\right) = 3\left(-\frac{11}{6}\right)^2 + 11\left(-\frac{11}{6}\right) + 5
\]

Calculating this step-by-step:

1. Calculate \( \left(-\frac{11}{6}\right)^2 \):

\[
\left(-\frac{11}{6}\right)^2 = \frac{121}{36}
\]

2. Substitute this into \( f\left(-\frac{11}{6}\right) \):

\[
f\left(-\frac{11}{6}\right) = 3\left(\frac{121}{36}\right) + 11\left(-\frac{11}{6}\right) + 5
\]

This simplifies to:

\[
f\left(-\frac{11}{6}\right) = \frac{363}{36} - \frac{121}{6} + 5
\]

Convert 5 to a fraction with a denominator of 36:

\[
5 = \frac{180}{36}
\]

Thus:

\[
f\left(-\frac{11}{6}\right) = \frac{363 - 726 + 180}{36} = \frac{-183}{36} \approx -5.08
\]

### Summary

- The x-intercepts are approximately: \( -1.55, -3.45 \)
- The vertex is approximately: \( \left(-1.83, -5.08\right) \)

So, the final results are:

**X-Intercepts:** \(-1.55, -3.45\)
**Vertex:** \((-1.83, -5.08)\)