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 \((4,0)\)
