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.

Answer:

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

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 [tex]a[/tex], [tex]b[/tex] and [tex]c[/tex].

Denominator

i) [tex](x+2)\cdot (x-1) = 0[/tex] Given

ii) [tex]x^{2} + x-2 = 0[/tex] Factorization

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

After a quick comparison, we conclude that [tex]a = 2[/tex] and [tex]b = -4[/tex]

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

Numerator

i) [tex]c\cdot x - 5\cdot x^{2} = 0[/tex] Given

ii) [tex]x \cdot (c-5\cdot x) = 0[/tex] Distributive Property

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

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

i) [tex]\left(x -\frac{c}{5}\right) = (x+2)[/tex] Given

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

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

The value of [tex]c[/tex] is -10.

c) We can rewrite the rational function as:

[tex]f(x) = \frac{(-5\cdot x)\cdot \left(x+2 \right)}{2\cdot (x+2)\cdot (x-1)}[/tex]

After eliminating the removable discontinuity, the function becomes:

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

At [tex]x = -2[/tex], we find that [tex]f(-2)[/tex] is:

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

[tex]f(-2) = -\frac{5}{3}[/tex]

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

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

[tex]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][/tex] Modulative Property

[tex]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][/tex] Existence of Multiplicative Inverse/Definition of Division

[tex]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)[/tex]   [tex]\frac{\frac{x}{y} }{\frac{w}{z} } = \frac{x\cdot z}{y\cdot w}[/tex]

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

[tex]y = -\frac{5}{2}[/tex] Limit properties/[tex]\lim_{x \to \pm \infty} \frac{1}{x^{n}} = 0[/tex], for [tex]n \geq 1[/tex]

The horizontal asymptote to the graph of f is [tex]y = -\frac{5}{2}[/tex].

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 [tex]x = -2, f(x) = -\frac{5}{3}[/tex].

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

-------------------------

Item a:

[tex]2(x - 1)(x - (-2)) = 2(x - 1)(x + 2) = 2(x^2 + x - 2) = 2x^2 + 2x - 4[/tex]

[tex]2x^2 + ax + b = 2x^2 - 2x - 4[/tex]

Thus a = 2, b = -4.

-------------------------

Item b:

[tex]-2c - 5(-2)^2 = 0[/tex]

[tex]-2c - 20 = 0[/tex]

[tex]2c = -20[/tex]

[tex]c = -\frac{20}{2}[/tex]

[tex]c = -10[/tex]

-------------------------

Item c:

With the coefficients, the function is:

[tex]f(x) = \frac{-10x - 5x^2}{2x^2 + 2x - 4} = \frac{-5x(x + 2)}{2(x - 1)(x + 2)} = -\frac{5x}{2(x - 1)}[/tex]

At x = -2:

[tex]-\frac{5(-2)}{2(-2 - 1)} = -\frac{-10}{-6} = -(\frac{5}{3}) = -\frac{5}{3}[/tex]

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

-------------------------

Item d:

The horizontal asymptote of a function is:

[tex]y = \lim_{x \rightarrow \infty} f(x)[/tex]

Thus:

[tex]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}[/tex]

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

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