Find the Holes in the Graph f(x)=(2x^2-2)/(x^2-4x+3)

Solution:

Step:1

Extract the common factor from the numerator $2x^2-2$.

Step:1.1

Take out the factor of $2$ from the expression $2x^2-2$.

Step:1.1.1

Extract the factor $2$ from $2x^2$.
$f(x)=\frac{2(x^2)-2}{x^2-4x+3}$

Step:1.1.2

Extract the factor $2$ from $-2$.
$f(x)=\frac{2(x^2)+2(-1)}{x^2-4x+3}$

Step:1.1.3

Pull out the factor $2$ from the entire expression $2(x^2)+2(-1)$.
$f(x)=\frac{2(x^2-1)}{x^2-4x+3}$

Step:1.2

Express $1$ as $1^2$.
$f(x)=\frac{2(x^2-1^2)}{x^2-4x+3}$

Step:1.3

Apply factoring techniques.

Step:1.3.1

Use the difference of squares formula, $a^2-b^2=(a+b)(a-b)$, where $a=x$ and $b=1$.
$f(x)=\frac{2((x+1)(x-1))}{x^2-4x+3}$

Step:1.3.2

Eliminate unnecessary parentheses.
$f(x)=\frac{2(x+1)(x-1)}{x^2-4x+3}$

Step:2

Factor the denominator $x^2-4x+3$ by the AC method.

Step:2.1

Identify two integers whose product is $c$ and sum is $b$ for the quadratic form $x^2+bx+c$. Here, we need integers with a product of $3$ and a sum of $-4$.
The integers are $-3$ and $-1$.

Step:2.2

Write the denominator in its factored form using the identified integers.
$f(x)=\frac{2(x+1)(x-1)}{(x-3)(x-1)}$

Step:3

Eliminate the common factor of $x-1$ from the numerator and denominator.

Step:3.1

Cancel out the common factor.
$f(x)=\frac{2(x+1)\cancel{(x-1)}}{(x-3)\cancel{(x-1)}}$

Step:3.2

Rewrite the simplified expression.
$f(x)=\frac{2(x+1)}{x-3}$

Step:4

Identify the factors in the denominator that were canceled to determine the holes in the graph.
The canceled factor is $x-1$.

Step:5

Find the coordinates of the holes by setting each canceled factor equal to $0$ and substituting back into the simplified function $\frac{2(x+1)}{x-3}$.

Step:5.1

Set the canceled factor $x-1$ equal to $0$.
$x-1=0$

Step:5.2

Solve for $x$ by adding $1$ to both sides.
$x=1$

Step:5.3

Substitute $x=1$ into the simplified function to find the $y$-coordinate of the hole.

Step:5.3.1

Replace $x$ with $1$ to find the $y$-coordinate.
$\frac{2(1+1)}{1-3}$

Step:5.3.2

Perform the arithmetic operations.

Step:5.3.2.1

Add $1$ and $1$.
$\frac{2\cdot 2}{1-3}$

Step:5.3.2.2

Subtract $3$ from $1$.
$\frac{2\cdot 2}{-2}$

Step:5.3.2.3

Multiply $2$ by $2$.
$\frac{4}{-2}$

Step:5.3.2.4

Divide $4$ by $-2$.
The result is $-2$.

Step:5.4

The coordinates of the holes are the points where the canceled factors equal $0$.
The hole is at $(1,-2)$.

Step:6

There are no further steps; the process is complete.

Knowledge Notes:

To find holes in the graph of a rational function, we must first factor both the numerator and the denominator. Holes occur at values of $x$ where a common factor in both the numerator and denominator is canceled out. These are the points where the function is not defined, despite the fact that the limit exists as $x$ approaches the hole's $x$-coordinate.

The key steps in this process include:

1. Factoring the numerator and denominator separately.

2. Identifying and canceling common factors.

3. Determining the $x$-coordinates of the holes by setting the canceled factors equal to zero.

4. Finding the $y$-coordinates of the holes by substituting the $x$-coordinates into the simplified function.

The difference of squares formula, $a^2-b^2=(a+b)(a-b)$, is a useful algebraic identity for factoring expressions like $x^2-1$. The AC method is a factoring technique used to factor quadratics of the form $a{x}^2+bx+c$, where $a$, $b$, and $c$ are integers. It involves finding two numbers that multiply to $ac$ and add to $b$, and then using these numbers to split the middle term and factor by grouping.