2 Sketch the graph of h(x) of [ (x^2+5 x+6)/(x+1) ]

To sketch the graph of \( h(x) = \frac{x^2 + 5x + 6}{x + 1} \), we need to analyze the function's behavior, including its domain, intercepts, asymptotes, and general shape.<br /><br />### Step-by-Step Analysis:<br /><br />1. **Factor the Numerator:**<br /> \[<br /> x^2 + 5x + 6 = (x + 2)(x + 3)<br /> \]<br /> So, the function simplifies to:<br /> \[<br /> h(x) = \frac{(x + 2)(x + 3)}{x + 1}<br /> \]<br /><br />2. **Domain:**<br /> The function is undefined where the denominator is zero. Therefore, the domain is:<br /> \[<br /> x \neq -1<br /> \]<br /><br />3. **Intercepts:**<br /> - **x-intercepts:** Set the numerator equal to zero:<br /> \[<br /> (x + 2)(x + 3) = 0 \implies x = -2 \text{ or } x = -3<br /> \]<br /> So, the x-intercepts are at \( x = -2 \) and \( x = -3 \).<br /><br /> - **y-intercept:** Set \( x = 0 \):<br /> \[<br /> h(0) = \frac{0^2 + 5 \cdot 0 + 6}{0 + 1} = \frac{6}{1} = 6<br /> \]<br /> So, the y-intercept is at \( y = 6 \).<br /><br />4. **Asymptotes:**<br /> - **Vertical Asymptote:** The function is undefined at \( x = -1 \), so there is a vertical asymptote at \( x = -1 \).<br /> - **Horizontal Asymptote:** For large values of \( |x| \), the term \( \frac{x^2}{x} \) dominates, so:<br /> \[<br /> \lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{x^2 + 5x + 6}{x + 1} = \lim_{x \to \infty} \frac{x^2}{x} = x<br /> \]<br /> However, since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, there is an oblique asymptote given by the quotient:<br /> \[<br /> y = x + 4<br /> \]<br /><br />5. **Behavior Near Asymptotes:**<br /> - As \( x \to -1^+ \), \( h(x) \to +\infty \).<br /> - As \( x \to -1^- \), \( h(x) \to -\infty \).<br /><br />6. **Plotting Points:**<br /> Choose a few points to plot the graph accurately:<br /> - \( x = -4 \):<br /> \[<br /> h(-4) = \frac{(-4)^2 + 5(-4) + 6}{-4 + 1} = \frac{16 - 20 + 6}{-3} = \frac{2}{-3} = -\frac{2}{3}<br /> \]<br /> - \( x = -2 \):<br /> \[<br /> h(-2) = \frac{(-2)^2 + 5(-2) + 6}{-2 + 1} = \frac{4 - 10 + 6}{-1} = \frac{0}{-1} = 0<br /> \]<br /> - \( x = 0 \):<br /> \[<br /> h(0) = 6<br /> \]<br /><br />### Graph Sketch:<br />1. Draw the vertical asymptote at \( x = -1 \).<br />2. Plot the x-intercepts at \( x = -2 \) and \( x = -3 \).<br />3. Plot the y-intercept at \( y = 6 \).<br />4. Draw the oblique asymptote \( y = x + 4 \).<br />5. Sketch the curve approaching the vertical asymptote as \( x \to -1 \) and the oblique asymptote as \( x \to \pm\infty \).<br /><br />Here is a rough sketch of the graph:<br /><br />```<br />y<br />|<br />| /<br />| /<br />| /<br />| /<br />| /<br />| /<br />| /<br />| /<br />| /<br />| / <br />|/_________________________ x<br />-3 -2 -1 0 1 2 3<br />```<br /><br />In this sketch:<br />- The vertical line at \( x = -1 \) represents the vertical asymptote.<br />- The line \( y = x + 4 \) represents the oblique asymptote.<br />- The points \( (-2,