Mathematics, 03.09.2020 09:01 FortniteB

Let f be the function defined by f(x)=cx−5x^2/2x^2+ax+b, where a, b, and c are constants. The graph of f has a vertical asymptote at x=1, and f has a removable discontinuity at x=−2. (a) Show that a=2 and b=−4. (b) Find the value of c. Justify your answer. (c) To make f continuous at x=−2, f(−2) should be defined as what value? Justify your answer. (d) Write an equation for the horizontal asymptote to the graph of f. Show the work that leads to your answer.

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Answer from: gmc2771

Using asymptote concepts, it is found that:

a) Building a quadratic equation with leading coefficient 2 and roots 1 and -2, it is found that a = 2, b = -4.

b) c = -10, since the discontinuity at x = -2 is removable, the numerator is 0 when x = -2.

c) Simplifying the function, it is found that at $x = -2, f(x) = -\frac{5}{3}$ .

d) The equation for the horizontal asymptote to the graph of f is $y = -\frac{5}{2}$

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Item a:

Vertical asymptote at x = 1

and discontinuity at x = -2

means that the the roots of the quadratic function at the denominator are x = 1

and

.The leading coefficient is given as 2, thus, we build the equation to find coefficients a and b.

2(x - 1)(x - (-2)) = 2(x - 1)(x + 2) = 2(x^2 + x - 2) = 2x^2 + 2x - 4

2x^2 + ax + b = 2x^2 - 2x - 4

Thus a = 2, b = -4.

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Item b:

Removable discontinuity at x = -2

means that the numerator when x = -2

is 0, thus:

-2c - 5(-2)^2 = 0

-2c - 20 = 0

2c = -20

$c = -\frac{20}{2}$

c = -10

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Item c:

With the coefficients, the function is:

$f(x) = \frac{-10x - 5x^2}{2x^2 + 2x - 4} = \frac{-5x(x + 2)}{2(x - 1)(x + 2)} = -\frac{5x}{2(x - 1)}$

At x = -2:

$-\frac{5(-2)}{2(-2 - 1)} = -\frac{-10}{-6} = -(\frac{5}{3}) = -\frac{5}{3}$

Thus, simplifying the function, it is found that at $x = -2, f(x) = -\frac{5}{3}$

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Item d:

The horizontal asymptote of a function is:

$y = \lim_{x \rightarrow \infty} f(x)$

Thus:

$y = \lim_{x \rightarrow \infty} \frac{-10x - 5x^2}{2x^2 + 2x - 4} = \lim_{x \rightarrow \infty} \frac{-5x^2}{2x^2} = \lim_{x \rightarrow \infty} -\frac{5}{2} = -\frac{5}{2}$

The equation for the horizontal asymptote to the graph of f is $y = -\frac{5}{2}$

A similar problem is given at link

Answer from: bg988763p7cl2d

a) a = 2 and b = -4 , b) c = -10 , c) $f(-2) = -\frac{5}{3}$ , d) $y = -\frac{5}{2}$ .

Step-by-step explanation:

a) After we read the statement carefully, we find that rational-polyomic function has the following characteristics:

1) A root of the polynomial at numerator is -2. (Removable discontinuity)

2) Roots of the polynomial at denominator are 1 and -2, respectively. (Vertical asymptote and removable discontinuity.

We analyze each polynomial by factorization and direct comparison to determine the values of , and .

Denominator

i) $(x+2)\cdot (x-1) = 0$ Given

ii) $x^{2} + x-2 = 0$ Factorization

iii) $2\cdot x^{2}+2\cdot x -4 = 0$ Compatibility with multiplication/Cancellative Property/Result

After a quick comparison, we conclude that a = 2 and b = -4

b) The numerator is analyzed by applying the same approached of the previous item:

Numerator

i) $c\cdot x - 5\cdot x^{2} = 0$ Given

ii) $x \cdot (c-5\cdot x) = 0$ Distributive Property

iii) $(-5\cdot x)\cdot \left(x-\frac{c}{5}\right)=0$ Distributive and Associative Properties/ $(-a)\cdot b = -a\cdot b$ /Result

As we know, this polynomial has x = -2 as one of its roots and therefore, the following identity must be met:

i) $\left(x -\frac{c}{5}\right) = (x+2)$ Given

ii) $\frac{c}{5} = -2$ Compatibility with addition/Modulative property/Existence of additive inverse.

iii) c = -10 Definition of division/Existence of multiplicative inverse/Compatibility with multiplication/Modulative property/Result

The value of is -10.

c) We can rewrite the rational function as:

$f(x) = \frac{(-5\cdot x)\cdot \left(x+2 \right)}{2\cdot (x+2)\cdot (x-1)}$

After eliminating the removable discontinuity, the function becomes:

$f(x) = -\frac{5}{2}\cdot \left(\frac{x}{x-1}\right)$

At x = -2 , we find that f(-2) is:

$f(-2) = -\frac{5}{2}\cdot \left[\frac{(-2)}{(-2)-1} \right]$

$f(-2) = -\frac{5}{3}$

d) The value of the horizontal asympote is equal to the limit of the rational function tending toward $\pm \infty$ . That is:

$y = \lim_{x \to \pm\infty} \frac{-10\cdot x-5\cdot x^{2}}{2\cdot x^{2}+2\cdot x -4}$ Given

$y = \lim_{x \to \infty} \left[\left(\frac{-10\cdot x-5\cdot x^{2}}{2\cdot x^{2}+2\cdot x-4}\right)\cdot 1\right]$ Modulative Property

$y = \lim_{x \to \infty} \left[\left(\frac{-10\cdot x-5\cdot x^{2}}{2\cdot x^{2}+2\cdot x-4}\right)\cdot \left(\frac{x^{2}}{x^{2}} \right)\right]$ Existence of Multiplicative Inverse/Definition of Division

$y = \lim_{x \to \pm \infty} \left(\frac{\frac{-10\cdot x-5\cdot x^{2}}{x^{2}} }{\frac{2\cdot x^{2}+2\cdot x -4}{x^{2}} } \right)$ $\frac{\frac{x}{y} }{\frac{w}{z} } = \frac{x\cdot z}{y\cdot w}$

$y = \lim_{x \to \pm \infty} \left(\frac{-\frac{10}{x}-5 }{2+\frac{2}{x}-\frac{4}{x^{2}} } \right)$ $\frac{x}{y} + \frac{z}{y} = \frac{x+z}{y}$ / $x^{m}\cdot x^{n} = x^{m+n}$