determine all the x - intercepts, asymptotes, and points of discontinuity of the given function. if there is…

determine all the x - intercepts, asymptotes, and points of discontinuity of the given function. if there is more than one value for any of them then separate them with a comma and no spaces and with smaller numbers going first. for example, -2,1,4. if there are no answers then place the word none in the corresponding answer box. make sure you write the point of discontinuity as an ordered pair such as (3,4) with no spaces. x - intercept(s) is/are vertical asymptote(s) is/are x = horizontal asymptote is y = point(s) of discontinuity is y=(x^2 - 3x - 4)/(x^2 - 3x + 2)

determine all the x - intercepts, asymptotes, and points of discontinuity of the given function. if there is more than one value for any of them then separate them with a comma and no spaces and with smaller numbers going first. for example, -2,1,4. if there are no answers then place the word none in the corresponding answer box. make sure you write the point of discontinuity as an ordered pair such as (3,4) with no spaces. x - intercept(s) is/are vertical asymptote(s) is/are x = horizontal asymptote is y = point(s) of discontinuity is y=(x^2 - 3x - 4)/(x^2 - 3x + 2)

Answer

Explanation:

Step1: Find x - intercepts

Set $y = 0$, then $\frac{x^{2}-3x - 4}{x^{2}-3x + 2}=0$. For a fraction to be zero, the numerator must be zero while the denominator is non - zero. Solve $x^{2}-3x - 4=0$. Factoring gives $(x - 4)(x+1)=0$, so $x = 4$ or $x=-1$. When $x = 4$, $x^{2}-3x + 2=16-12 + 2=6\neq0$ and when $x=-1$, $x^{2}-3x + 2=1 + 3+2=6\neq0$.

Step2: Find vertical asymptotes

Set the denominator equal to zero. Solve $x^{2}-3x + 2=0$. Factoring gives $(x - 2)(x - 1)=0$. So $x = 1$ and $x = 2$ are the vertical asymptotes.

Step3: Find horizontal asymptote

Since the degree of the numerator and the denominator are the same (both degree 2), the horizontal asymptote is $y=\frac{a}{b}$, where $a$ and $b$ are the leading coefficients of the numerator and denominator. Here $a = 1$ and $b = 1$, so $y = 1$.

Step4: Find points of discontinuity

The function is discontinuous at the values of $x$ that make the denominator zero, which are $x = 1$ and $x = 2$. But since the numerator is non - zero at these points, these are vertical asymptotes, not removable discontinuities. So there are no points of discontinuity in the form of an ordered pair.

Answer:

x - intercept(s): $-1,4$ vertical asymptote(s) $x = 1,2$ horizontal asymptote $y = 1$ point(s) of discontinuity: none