Let's solve step-by-step for the function \( y = \frac{x^2}{x+1} \): --- ### 1. **Domain** The function is undefined when the denominator is zero: \[ x + 1 = 0 \implies x = -1 \] **Domain:** \( (-\infty, -1) \cup (-1, \infty) \) --- ### 2. **Vertical Asymptote** Vertical asymptotes occur where the denominator is zero (and numerator is nonzero): \[ x = -1 \] --- ### 3. **Horizontal/Oblique Asymptote** For large \( |x| \), compare degrees (numerator degree is 2, denominator is 1): - Since the degree of the numerator is one more than the denominator, there is an **oblique (slant) asymptote**. - Perform polynomial division: \( x^2 \div (x + 1) \): \[ \frac{x^2}{x+1} = x - 1 + \frac{1}{x+1} \] So the **slant asymptote** is \( y = x - 1 \). --- ### 4. **Intercepts** - **x-intercept:** Set \( y = 0 \implies x^2 = 0 \implies x = 0 \) - So, \( (0, 0) \) is the x-intercept. - **y-intercept:** Set \( x = 0 \implies y = 0 \) - So, \( (0, 0) \) is also the y-intercept. --- ### 5. **Local Maxima/Minima (Critical Points)** Find where \( y' = 0 \): \[ y = \frac{x^2}{x+1} \] Use quotient rule: \[ y' = \frac{(x+1) \cdot 2x - x^2 \cdot 1}{(x+1)^2} = \frac{2x(x+1) - x^2}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2} = \frac{x(x+2)}{(x+1)^2} \] Set numerator to zero: \[ x(x+2) = 0 \implies x = 0 \text{ or } x = -2 \] - At \( x = 0 \): \( y = 0 \) - At \( x = -2 \): \( y = \frac{(-2)^2}{-2+1} = \frac{4}{-1} = -4 \) --- ### 6. **Increasing/Decreasing Intervals** Check sign of \( y' \): - \( x(x+2) > 0 \) when \( x < -2 \) or \( x > 0 \) (**increasing**) - \( x(x+2) < 0 \) when \( -2 < x < 0 \) (**decreasing**) But remember \( x = -1 \) is not in the domain. So: - **Increasing:** \( (-\infty, -2) \cup (0, -1) \cup (0, \infty) \) - **Decreasing:** \( (-2, -1) \cup (-1, 0) \) --- ### 7. **Second Derivative (Concavity & Inflection Point)** Differentiate \( y' \): \[ y' = \frac{x(x+2)}{(x+1)^2} \] Let \( u = x(x+2),\ v = (x+1)^2 \): \[ y'' = \frac{v \cdot u' - u \cdot v'}{v^2} \] \[ u' = 2x+2 \] \[ v' = 2(x+1) \] \[ y'' = \frac{(x+1)^2(2x+2) - x(x+2) \cdot 2(x+1)}{(x+1)^4} \] Factor \( (x+1) \): \[ = \frac{(x+1)[(x+1)(2x+2) - 2x(x+2)]}{(x+1)^4} \] \[ (x+1)(2x+2) = 2x^2 + 2x + 2x + 2 = 2x^2 + 4x + 2 \] \[ 2x(x+2) = 2x^2 + 4x \] \[ = \frac{(x+1)[2x^2 + 4x + 2 - (2x^2 + 4x)]}{(x+1)^4} = \frac{(x+1)[2]}{(x+1)^4} = \frac{2(x+1)}{(x+1)^4} = \frac{2}{(x+1)^3} \] Set \( y'' = 0 \): \[ \frac{2}{(x+1)^3} = 0 \] But this never equals zero. So **no inflection point**. --- ### 8. **Range** - As \( x \to \infty \), \( y \to x-1 \to \infty \) - As \( x \to -\infty \), \( y \to x-1 \to -\infty \) - At \( x = -2 \), \( y = -4 \) (local minimum) - At \( x = 0 \), \( y = 0 \) (local maximum) - Function can take all real values except possibly between the local min/max, but because the function is continuous on both sides and goes to \(\pm\infty\), **the range is \( (-\infty, \infty) \)**. --- ### 9. **Summary Table** | Property | Value/Equation | |---------------------------|------------------------------| | **Domain** | \( (-\infty, -1) \cup (-1, \infty) \) | | **Vertical Asymptote** | \( x = -1 \) | | **Slant Asymptote** | \( y = x - 1 \) | | **x- and y-Intercept** | \( (0,0) \) | | **Local Minimum** | \( x = -2,\, y = -4 \) | | **Local Maximum** | \( x = 0,\, y = 0 \) | | **Increasing** | \( (-\infty, -2) \cup (0, -1) \cup (0, \infty) \) | | **Decreasing** | \( (-2, -1) \cup (-1, 0) \) | | **Inflection Points** | None | | **Range** | \( (-\infty, \infty) \) | --- ### 10. **Graph** - Vertical asymptote at \( x = -1 \) - Slant asymptote \( y = x-1 \) - Local minimum at \( (-2, -4) \), local maximum at \( (0,0) \) --- **Final answer**: All the above properties, and you can sketch the graph using this information. If you need the graph plotted, let me know!