Question

Let f be the function defined by f(x)=ax^2+bx+2/2x^2−8, where a and b are constants. The graph of f has a horizontal asymptote at y=3, and f has a removable discontinuity at x=2. What is the value of b?

Answer 1

Answer:

The value of b is -13

Step-by-step explanation:

* Lets explain how to find a and b

- In any rational function f(x), if the degree of the denominator =

 the degree of the numerator, then there is a horizontal asymptote

 at y = coefficient of higher x of the numerator ÷ coefficient of higher

 x of the denominator

- Removable discontinuity means the numerator and denominator of

 the rational function have common factor will reduce when we

 simplify the fraction

- Discontinuity occurs when a number is both a zero of the numerator

 and denominator

 

* Lets solve the problem

∵ $f(x)=\frac{ax^{2}+bx+2}{2x^{2}-8}$

∵ The degree of numerator and denominator is 2

∵ The coefficient of x² in the numerator is a

∵ The coefficient of x² in the denominator is 2

∴ The horizontal asymptote is at $y=\frac{a}{2}$

∵ The graph of f has a horizontal asymptote at y = 3

∴ $\frac{a}{2}=3$

- By using cross multiplication

∴ a = 6

∴ $f(x)=\frac{6x^{2}+bx+2}{2x^{2}-8}$

∵ The function f has a removable discontinuity at x = 2

∵ Discontinuity occurs when a number is both a zero of the numerator

  and denominator

∴ 2 is a zero of the numerator

∴ 6x² + bx + 2 = 0 at x = 2

- Substitute x by 2 to find b

∴ 6(2)² + b(2) + 2 = 0

∴ 6(4) + 2b + 2 = 0

∴ 24 + 2b + 2 = 0

∴ 26 + 2b = 0

- Subtract 26 from both sides

∴ 2b = -26

- Divide both sides by 2

∴ b = -13