Details the method for identifying and locating the coordinates of holes in the graph of a rational function, which occur when a factor in the numerator and denominator cancels out.
What is a hole in the graph of a rational function?
A hole is a single point at which the function is undefined. It's also called a 'removable discontinuity' because the function can be redefined to 'plug' the hole.
How do you algebraically identify a factor that corresponds to a hole?
A hole occurs when a factor, like \((x-c)\), appears in both the numerator and the denominator of the rational function. This common factor can be canceled out.
How do you find the x-coordinate of a hole?
Set the common factor that was canceled from the numerator and denominator equal to zero and solve for \(x\).
After finding the x-coordinate of a hole, how do you find its y-coordinate?
Plug the x-coordinate of the hole into the *simplified* version of the rational function (after the common factor has been canceled) and solve for \(y\).
Find the coordinates of the hole for the function \(f(x) = \frac{x^2 - 25}{x - 5}\).
1. Factor: \(f(x) = \frac{(x-5)(x+5)}{x-5}\).
2. Cancel \((x-5)\). The hole is at \(x=5\).
3. Use the simplified function \(f_{simp}(x) = x+5\).
4. Plug in \(x=5\): \(y = 5+5=10\). The hole is at \((5, 10)\).
Identify all discontinuities in \(g(x) = \frac{x-1}{x^2 - 1}\) and classify them.
Factor the denominator: \(g(x) = \frac{x-1}{(x-1)(x+1)}\).
- The factor \((x-1)\) cancels, so there is a hole at \(x=1\).
- The factor \((x+1)\) remains in the denominator, so there is a vertical asymptote at \(x=-1\).
Find the coordinates of the hole for \(g(x) = \frac{x-1}{x^2-1}\).
The hole is at \(x=1\). The simplified function is \(g_{simp}(x) = \frac{1}{x+1}\). Plug in \(x=1\) to get \(y = \frac{1}{1+1} = \frac{1}{2}\). The hole is at \((1, 1/2)\).
What is the visual representation of a hole on a graph?
A hole is drawn as a small open circle on the graph at the point's coordinates, indicating that the function is defined all around that point but not at the point itself.
