Question

Identify all the vertical asymptotes for the function f(x)= 2x^2-11x-6/2x^3-15x^2+18x

Answer 1

To find the vertical asymptotes of the function, set the denominator equal to zero and solve for x. The values of x that make the denominator equal to zero are x = 0, x = 3/2, and x = 6.

Step-by-step explanation:

To identify the vertical asymptotes of the function f(x) = (2x^2 - 11x - 6) / (2x^3 - 15x^2 + 18x), we need to find the values of x that make the denominator equal to zero.

To find these values, set the denominator equal to zero and solve for x:

2x^3 - 15x^2 + 18x = 0

Now, factor out an x:

x(2x^2 - 15x + 18) = 0

Factor the quadratic equation:

(2x - 3)(x - 6) = 0

So, the values of x that make the denominator equal to zero are x = 0, x = 3/2, and x = 6. These are the vertical asymptotes of the function f(x).

Answer 2

The function f(x) has vertical asymptotes at:

x = 0,

x = 6, and

x = 1.5.

Step-by-step explanation:

Solving 2x^3 - 15x^2 + 18x = 0:

Factoring the polynomial:

We notice 2x appears in all terms, so we factor it out:

2x(x^2 - 7.5x + 9)

Factoring the remaining polynomial:

The remaining polynomial (x^2 - 7.5x + 9) looks factorable, so we try grouping:

(x^2 - 6x) - (1.5x - 9)

We find a common factor of (x - 6) in both terms:

(x - 6)(x - 1.5)

Setting each factor equal to zero and solving:

If either factor equals zero, the original equation is satisfied. So we solve:

x - 6 = 0 --> x = 6

x - 1.5 = 0 --> x = 1.5

Therefore, the solutions are:

x = 0, x = 6, and x = 1.5