Problem 59
Question
Sketch a graph of the rational function and find all intercepts and slant asymptotes. Indicate all asymptotes onthe graph. $$h(x)=\frac{x^{2}+x+1}{x-1}$$
Step-by-Step Solution
Verified Answer
This function has a y-intercept at (0,-1) and no x-intercept. The vertical asymptote is \(x = 1\) and the slant asymptote is \(y=x+2\).
1Step 1: Find x-intercepts
To solve this step, make the rational function equal to zero and solve for x. However, the numerator \(x^{2} +x+1=0\) has complex roots, so the graph does not cross the x-axis. Therefore, there's no x-intercept.
2Step 2: Find y-intercept
The y-intercept occurs where \(x = 0\). Substitution \(x = 0\) to function, you find \(h(0) = -1\). So the y-intercept is at point (0,-1).
3Step 3: Determine Asymptotes
First, the vertical asymptote is given by the zeros of the denominator \(x-1=0\). So the vertical asymptote is \(x=1\). Since the degree of the numerator is greater than the degree of the denominator, there is a slant asymptote. To find the equation for the slant asymptote, perform polynomial division: \(x^2 + x + 1\) divided by \(x -1\). The result is \(x+2\) with reminder \(3\). So, the slant (or oblique) asymptote is \(y = x + 2\).
4Step 4: Sketch the Function
Once all intercepts and asymptotes have been found, plot them on a graph. Draw the asymptotes using dashed lines at \(x=1\) and \(y = x + 2\). The graph crosses the y-axis at (0,-1). The graph approaches and follows the asymptotes.
Key Concepts
AsymptotesX-interceptsY-interceptsGraphing Functions
Asymptotes
Rational functions can exhibit behavior where they approach certain lines without actually crossing them—these lines are called asymptotes. For the function \( h(x) = \frac{x^2 + x + 1}{x - 1} \), there are two kinds of asymptotes: vertical and slant.
Vertical asymptotes occur where the denominator of a function is zero, leading to undefined values for \( h(x) \). In our case, setting the denominator \( x - 1 = 0 \) gives \( x = 1 \) as the vertical asymptote.
Slant asymptotes show up when the degree of the numerator is greater than the degree of the denominator. To find it, perform polynomial division on \( x^2 + x + 1 \) by \( x - 1 \). This calculates to \( x + 2 \), giving the slant asymptote as \( y = x + 2 \). Slant asymptotes represent the curve's long-term behavior, drawing closer as \( x \) becomes very large or small.
Vertical asymptotes occur where the denominator of a function is zero, leading to undefined values for \( h(x) \). In our case, setting the denominator \( x - 1 = 0 \) gives \( x = 1 \) as the vertical asymptote.
Slant asymptotes show up when the degree of the numerator is greater than the degree of the denominator. To find it, perform polynomial division on \( x^2 + x + 1 \) by \( x - 1 \). This calculates to \( x + 2 \), giving the slant asymptote as \( y = x + 2 \). Slant asymptotes represent the curve's long-term behavior, drawing closer as \( x \) becomes very large or small.
- Vertical asymptote: \( x = 1 \)
- Slant asymptote: \( y = x + 2 \)
X-intercepts
X-intercepts of a function occur where the graph crosses the x-axis. It's where the output \( h(x) \) equals zero. For rational functions, this is found by setting the numerator to zero and solving for \( x \).
In our function \( h(x) = \frac{x^2 + x + 1}{x - 1} \), the numerator, \( x^2 + x + 1 \), results in complex roots. This means there are no real solutions and thus, no x-intercepts for this graph.
Understanding x-intercepts is crucial, as they highlight points where outputs flip from positive to negative. The absence of x-intercepts implies the graph does not touch or pass through the x-axis.
In our function \( h(x) = \frac{x^2 + x + 1}{x - 1} \), the numerator, \( x^2 + x + 1 \), results in complex roots. This means there are no real solutions and thus, no x-intercepts for this graph.
Understanding x-intercepts is crucial, as they highlight points where outputs flip from positive to negative. The absence of x-intercepts implies the graph does not touch or pass through the x-axis.
- No x-intercepts due to complex roots
Y-intercepts
Finding the y-intercept of a function involves determining where the graph crosses the y-axis. This happens when \( x \) is zero.
By substituting \( x = 0 \) into \( h(x) = \frac{x^2 + x + 1}{x - 1} \), you get \( h(0) = \frac{0^2 + 0 + 1}{0 - 1} = -1 \). Thus, the y-intercept for this graph is the point \( (0, -1) \).
Y-intercepts are vital for sketching functions, providing a visual anchor to help perfect the curve's form. You can always test this point in your graph to ensure accuracy during plotting.
By substituting \( x = 0 \) into \( h(x) = \frac{x^2 + x + 1}{x - 1} \), you get \( h(0) = \frac{0^2 + 0 + 1}{0 - 1} = -1 \). Thus, the y-intercept for this graph is the point \( (0, -1) \).
Y-intercepts are vital for sketching functions, providing a visual anchor to help perfect the curve's form. You can always test this point in your graph to ensure accuracy during plotting.
- Y-intercept: \( (0, -1) \)
Graphing Functions
Graphing rational functions means deploying both mathematical calculations and visual judgment. Begin by identifying key elements like intercepts and asymptotes. These guide how the curve will appear.
For our function \( h(x) = \frac{x^2 + x + 1}{x - 1} \):
Analyzing these elements can help you to anticipate where the graph will curve and bend. Rational function graphs usually exhibit symmetry and consistency along these marked guidelines.
For our function \( h(x) = \frac{x^2 + x + 1}{x - 1} \):
- Draw the vertical asymptote \( x = 1 \) and slant asymptote \( y = x + 2 \) using dashed lines.
- Mark the y-intercept at \( (0, -1) \) on the graph.
Analyzing these elements can help you to anticipate where the graph will curve and bend. Rational function graphs usually exhibit symmetry and consistency along these marked guidelines.
Other exercises in this chapter
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