Question

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

Answer 1

3x^2-11x-4 = (3x+1)(x-4)

The negative x-intercept is at (-1/3,0)
The positive x-intercept is at (4,0)

Answer 2

we know that  

The x-intercept is the value of x when the value of the function is equal to zero.

in this problem  we have

$f(x)=3x^{2}-11x-4$

To find the x-intercepts equate the function to zero

$3x^{2}-11x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$3x^{2}-11x=4$

Factor the leading coefficient

$3(x^{2}-(11x/3))=4$

Complete the square. Remember to balance the equation by adding the same constants to each side

$3(x^{2}-(11x/3)+(121/36))=4+(121/12)$

$3(x^{2}-(11x/3)+(121/36))=(169/12)$

$(x^{2}-(11x/3)+(121/36))=(169/36)$

Rewrite as perfect squares

$(x-(11/6))^{2}=(169/36)$

square root both sides

$(x-(11/6))=(+/-)(13/6)$

$x=(11/6)(+/-)(13/6)$

$x=(11/6)+(13/6)=24/6=4$

$x=(11/6)-(13/6)=-2/6=-1/3$

therefore

the answer is

The negative x-intercept is at $(-1/3,0)$

The positive x-intercept is at $(4,0)$