Mathematics, 12.09.2019 08:10 aaburto1515

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?

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Answer from: kathleendthomas

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

Answer from: 24swimdylanoh

b = -13

Step-by-step explanation:

Given :

$\rm f(x) = \dfrac{ax^2+bx+2}{2x^2-8}$ ----- (1)

The graph of f has a horizontal asymptote at y = 3 and f has a removable discontinuity at x = 2.

Solution :

Horizontal assymptote is

$y = \dfrac{a}{2}$

because in equation (1) coefficient of x^2 is 'a' is in numerator and '2' is in denomenator.

So,

a = 6

Now equation (1) becomes,

$\rm f(x) = \dfrac{6x^2+bx+2}{2x^2-8}$ ------ (2)

Given that discontinuity at x = 2, so denomenator of equation (2) is zero at x = 2. Therefore

$\rm 6x^2 + bx + 2 = 0$

6(2^2) + 2b +2 = 0

-26 = 2b

b = -13

For more information, refer the link given below

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Answer from: Quest

step-by-step explanation:

Answer from: Quest

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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...