Question

Where does f(x) = 3x2 – 11x – 4 intersect the x-axis?





The negative x-intercept is at (
, 0).

The positive x-intercept is at (
, 0).

Answer 1

To find the x-intercepts of the function \( f(x) = 3x^2 - 11x - 4 \), we need to determine where \( f(x) = 0 \). This means we need to solve the equation:

\[
3x^2 - 11x - 4 = 0
\]

We can solve this quadratic equation using the quadratic formula:

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

Here, \( a = 3 \), \( b = -11 \), and \( c = -4 \).

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

\[
b^2 - 4ac = (-11)^2 - 4 \cdot 3 \cdot (-4) = 121 + 48 = 169
\]

2. Now, substitute into the quadratic formula:

\[
x = \frac{-(-11) \pm \sqrt{169}}{2 \cdot 3}
\]

\[
x = \frac{11 \pm 13}{6}
\]

This gives us two solutions:

- \( x = \frac{11 + 13}{6} = \frac{24}{6} = 4 \)
- \( x = \frac{11 - 13}{6} = \frac{-2}{6} = -\frac{1}{3} \)

Thus, the x-intercepts are:
- The negative x-intercept is at \( \left(-\frac{1}{3}, 0\right) \).
- The positive x-intercept is at \( (4, 0) \).

So, the final answers are:
- The negative x-intercept is at \( \left(-\frac{1}{3}, 0\right) \).
- The positive x-intercept is at \( (4, 0) \).