Question

Let f be the function defined by f(x) = bx² - x/(3b) + x²/b²What is the value of b such that

f(x) has a vertical asymptote at x=2?

A) b=1
B) b=2
C) b=3
D) b=4

Answer 1

To find the value of b such that f(x) has a vertical asymptote at x=2, set the denominator equal to zero and solve for b. The value of b that satisfies this condition is b = -2.

Step-by-step explanation:

To find the value of b such that f(x) has a vertical asymptote at x=2, we need to determine the value of b that makes the denominator of the function equal to 0 at x=2.

Let's set the denominator equal to 0:

x/(3b) + x²/b² = 0

Multiplying through by b² to get rid of the denominators:

x·b + x² = 0

Factoring out an x:

x(b + x) = 0

To have a vertical asymptote at x=2, we need the factor (b + x) to equal 0 at x=2. Therefore, (b + 2) = 0, which implies b = -2.

So, the value of b such that f(x) has a vertical asymptote at x=2 is b = -2.