Let's solve the problem of graphing the rational function [tex]\( h(x) = \frac{-3x + 12}{x^2 - 5x + 4} \)[/tex] step-by-step:
1. Factor the Rational Function:
First, factor both the numerator and the denominator of the given function [tex]\( h(x) \)[/tex]:
- Numerator: [tex]\(-3x + 12 = -3(x - 4)\)[/tex]
- Denominator: [tex]\(x^2 - 5x + 4 = (x - 4)(x - 1)\)[/tex]
Thus, the function can be rewritten as:
[tex]\[
h(x) = \frac{-3(x - 4)}{(x - 4)(x - 1)}
\][/tex]
2. Identify Any Holes:
Since [tex]\(x - 4\)[/tex] appears in both the numerator and the denominator, we have a hole at [tex]\(x = 4\)[/tex]. To find the [tex]\(y\)[/tex]-value of the hole, we simplify the function:
[tex]\[
h(x) = \frac{-3(x - 4)}{(x - 4)(x - 1)} = \frac{-3}{x - 1}, \quad \text{for } x \neq 4.
\][/tex]
Then, substitute [tex]\(x = 4\)[/tex] into the simplified function to find the y-coordinate of the hole:
[tex]\[
h(4) = \frac{-3}{4 - 1} = \frac{-3}{3} = -1.
\][/tex]
Therefore, there is a hole at [tex]\((4, -1)\)[/tex].
3. Vertical Asymptotes:
The vertical asymptotes occur where the denominator is zero and the numerator is non-zero. In this case, the denominator [tex]\(x - 1\)[/tex] causes a vertical asymptote at [tex]\(x = 1\)[/tex].
4. Horizontal Asymptote:
For a rational function where the degree of the numerator and the degree of the denominator are the same, the horizontal asymptote is the ratio of the leading coefficients. In this function, the degrees are both 1. The leading coefficients are [tex]\(-3\)[/tex] (numerator) and [tex]\(1\)[/tex] (denominator), so the horizontal asymptote is:
[tex]\[
y = \frac{-3}{1} = -3.
\][/tex]
5. Plot Points:
Choose values for [tex]\( x \)[/tex] to find corresponding points on the graph. Avoid the values where the function is undefined (i.e., at the vertical asymptote [tex]\(x = 1\)[/tex] and hole [tex]\(x = 4\)[/tex]):
- For [tex]\( x = 0 \)[/tex]:
[tex]\[
h(0) = \frac{-3}{0 - 1} = \frac{-3}{-1} = 3.
\][/tex]
This gives the point [tex]\((0, 3)\)[/tex].
- For [tex]\( x = 2 \)[/tex]:
[tex]\[
h(2) = \frac{-3}{2 - 1} = \frac{-3}{1} = -3.
\][/tex]
This gives the point [tex]\((2, -3)\)[/tex].
6. Graphing the Function:
- Draw the vertical asymptote [tex]\(x = 1\)[/tex].
- Draw the horizontal asymptote [tex]\(y = -3\)[/tex].
- Plot the points found: [tex]\((0, 3)\)[/tex] and [tex]\((2, -3)\)[/tex].
- Draw a hollow dot (representing the hole) at [tex]\((4, -1)\)[/tex].
Connect the plotted points in a manner that the behavior of the graph aligns with asymptotes and the hole. The function should approach but never touch the vertical and horizontal asymptotes. Here's a rough sketch of the graph based on these descriptions.
If you were to create this graph, ensure it accurately depicts all of these characteristics.
