Find the Holes in the Graph f(x)=(2x^2-2)/(x^2-4x+3)
The given problem is asking to identify the locations (if any) of holes in the graph of the function f(x) = (2x^2 - 2)/(x^2 - 4x + 3). Holes in the graph of a function occur at points where the function is not defined due to a cancellation of factors in the numerator and denominator in its rational expression, which results in an indeterminate form at a certain x-value. To find the holes, you would factor both the numerator and the denominator of the function and look for common factors. If there are common factors, the values of x that make these factors equal to zero are the x-coordinates of the holes. However, to complete the description of the hole(s), you also have to determine the y-coordinate(s) by finding the limit of the function as x approaches the value at which the hole occurs, after the common factor has been canceled out.
$f \left(\right. x \left.\right) = \frac{2 x^{2} - 2}{x^{2} - 4 x + 3}$
Answer
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:
Factoring the numerator and denominator separately.
Identifying and canceling common factors.
Determining the $x$-coordinates of the holes by setting the canceled factors equal to zero.
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 $ax^2 + bx + c$, 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.
