Answer:
The non-negative zero of the function f(x) is x = 2.
Step-by-step explanation:
For a given function f(x), the "zeros" of the function are the values of x such that:
f(x) = 0
In this case, we have the function:
f(x) = 6*x^2 - 9*x - 6
If we want to find the zeros of this function, we need to solve:
f(x) = 0 = 6*x^2 - 9*x - 6
To solve this, we can use the Bhaskara's formula, which says that for a general quadratic equation:
0 = a*x^2 + b*x + c
The zeros are:
[tex]x = \frac{-b \pm \sqrt{b^2 - 4*a*c} }{2*a}[/tex]
In this case our equation is:
0 = 6*x^2 - 9*x - 6
then, in the above notation, we have:
a = 6
b = -9
c = -6
Replacing these in our general formula, we get:
[tex]x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4*6*(-6)} }{2*6} = \frac{9 \pm 15 }{12}[/tex]
Then we have two zeros:
x = (9 + 15)/12 = 24/12 = 2
x = (9 - 15)/12 = -6/12 = -1/2
But we want only the non-negative, so we can discard the second one.
Concluding, the non-negative zero of the function f(x) is x = 2.
