Question

1. sketch the graph. the factors have been given. just simplify.

$f(x)=\frac{x^{2}-25}{x^{2}-4x - 5}=$

coordinates of hole:

equation of vertical asymptote:

equation of horizontal or oblique asymptote:

domain: range:

x - intercept:

y - intercept:

does the graph cross the horizontal asymptote? show your work.

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-25=(x + 5)(x - 5)$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The denominator $x^{2}-4x - 5=(x-5)(x + 1)$ by factoring the quadratic $ax^{2}+bx + c$ where $a = 1$, $b=-4$, $c=-5$ and finding two numbers that multiply to $ac=-5$ and add to $b=-4$ (the numbers are $-5$ and $1$). So $f(x)=\frac{(x + 5)(x - 5)}{(x-5)(x + 1)}$.

Step2: Simplify the function

Cancel out the common factor $(x - 5)$ (for $x
eq5$). The simplified function is $f(x)=\frac{x + 5}{x + 1},x
eq5$.

Step3: Find the coordinates of the hole

The hole occurs at the value of $x$ that makes the original non - simplified denominator zero and was canceled out. Set $x-5 = 0$, so $x = 5$. Substitute $x = 5$ into the simplified function $y=\frac{5 + 5}{5+1}=\frac{10}{6}=\frac{5}{3}$. The coordinates of the hole are $(5,\frac{5}{3})$.

Step4: Find the vertical asymptote

Set the denominator of the simplified function equal to zero. $x+1=0$, so $x=-1$ is the equation of the vertical asymptote.

Step5: Find the horizontal or oblique asymptote

Since the degree of the numerator and the denominator of the simplified function are the same (both degree 1), the horizontal asymptote is $y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$. The leading coefficient of the numerator is 1 and of the denominator is 1, so $y = 1$ is the horizontal asymptote.

Step6: Find the domain

The domain is all real numbers except the values that make the denominator zero. So the domain is $\{x|x
eq-1,x
eq5\}$ or $(-\infty,-1)\cup(-1,5)\cup(5,\infty)$.

Step7: Find the range

The range is all real numbers except the value of the horizontal asymptote, so the range is $\{y|y
eq1\}$ or $(-\infty,1)\cup(1,\infty)$.

Step8: Find the x - intercept

Set $y = 0$. Then $\frac{x + 5}{x + 1}=0$, so $x+5=0$ and $x=-5$. The x - intercept is $x=-5$.

Step9: Find the y - intercept

Set $x = 0$. Then $y=\frac{0 + 5}{0+1}=5$. The y - intercept is $y = 5$.

Step10: Check if the graph crosses the horizontal asymptote

Set $y = 1$ (the horizontal asymptote). Then $\frac{x + 5}{x + 1}=1$. Cross - multiply: $x + 5=x + 1$, which simplifies to $5=1$, a contradiction. So the graph does not cross the horizontal asymptote.

Answer:

Coordinates of hole: $(5,\frac{5}{3})$
Equation of vertical asymptote: $x=-1$
Equation of horizontal asymptote: $y = 1$
Domain: $\{x|x
eq-1,x
eq5\}$ or $(-\infty,-1)\cup(-1,5)\cup(5,\infty)$
Range: $\{y|y
eq1\}$ or $(-\infty,1)\cup(1,\infty)$
X - intercept: $x=-5$
Y - intercept: $y = 5$
Does the graph cross the horizontal asymptote: No