Let’s go step-by-step

Let’s go step-by-step. --- **We are given:** \[ f(x) = \frac{x^2 - 1}{x^2 + x - 6} \] --- ### **Step 1: Factor numerator and denominator** Numerator: \[ x^2 - 1 = (x - 1)(x + 1) \] Denominator: \[ x^2 + x - 6 \] We look for two numbers that multiply to \(-6\) and add to \(1\): \(3\) and \(-2\). So: \[ x^2 + x - 6 = (x + 3)(x - 2) \] Thus: \[ f(x) = \frac{(x - 1)(x + 1)}{(x + 3)(x - 2)} \] --- ### **Step 2: Domain restrictions** Denominator zero when \(x = -3\) or \(x = 2\). So domain: all real \(x\) except \(x = -3, 2\). --- ### **Step 3: Holes (removable discontinuities)** Holes occur if a factor cancels from numerator and denominator. No common factors in numerator and denominator (since \(x-1\), \(x+1\), \(x+3\), \(x-2\) are all distinct). So **no holes**. --- ### **Step 4: Vertical asymptotes** Vertical asymptotes at \(x = -3\) and \(x = 2\) (since denominator zero, numerator nonzero at these points). Check: At \(x = -3\): numerator \(= (-4)(-2) = 8 \neq 0\) → VA at \(x = -3\). At \(x = 2\): numerator \(= (1)(3) = 3 \neq 0\) → VA at \(x = 2\). --- ### **Step 5: Horizontal asymptote** Degree of numerator = 2, degree of denominator = 2. Leading coefficient of numerator: \(1\) Leading coefficient of denominator: \(1\) So horizontal asymptote: \[ y = \frac{1}{1} = 1 \] --- ### **Step 6: x-intercepts** Set numerator = 0: \[ (x - 1)(x + 1) = 0 \implies x = 1, \quad x = -1 \] Both are in domain, so x-intercepts: \((-1, 0)\) and \((1, 0)\). --- ### **Step 7: y-intercept** Set \(x = 0\): \[ f(0) = \frac{-1}{-6} = \frac{1}{6} \] So y-intercept: \((0, \frac{1}{6})\). --- ### **Step 8: Sign chart and behavior near asymptotes** We have vertical asymptotes at \(x = -3\) and \(x = 2\), and roots at \(x = -1, 1\). Intervals: \((-\infty, -3)\), \((-3, -1)\), \((-1, 1)\), \((1, 2)\), \((2, \infty)\). Test points: - \(x = -4\): \(f(-4) = \frac{( -5)(-3)}{(-1)(-6)} = \frac{15}{6} > 0\) → positive. - \(x = -2\): \(f(-2) = \frac{(-3)(-1)}{(1)(-4)} = \frac{3}{-4} < 0\) → negative. - \(x = 0\): \(f(0) = 1/6 > 0\) → positive. - \(x = 1.5\): \(f(1.5) = \frac{(0.5)(2.5)}{(4.5)(-0.5)}\) → denominator negative, numerator positive → negative. - \(x = 3\): \(f(3) = \frac{(2)(4)}{(6)(1)} = 8/6 > 0\) → positive. So sign: \(+\) on \((-\infty, -3)\), \(-\) on \((-3, -1)\), \(+\) on \((-1, 1)\), \(-\) on \((1, 2)\), \(+\) on \((2, \infty)\). --- ### **Step 9: Behavior near vertical asymptotes** As \(x \to -3^-\) (from left): Numerator → \((-4)(-2) = 8\) (positive), denominator: \((x+3)(x-2)\), \(x \to -3^-\): \(x+3 \to 0^-\), \(x-2 \to -5\), product → \(0^+ \times (-5) = 0^-\)? Wait carefully: Let’s take \(x = -3.1\): \(x+3 = -0.1\) (negative), \(x-2 = -5.1\) (negative), product = positive. So denominator positive? Let’s check: \(x < -3\) → \(x+3 < 0\), \(x-2 < 0\) → product positive. So denominator positive, numerator positive → \(f(x) \to +\infty\) as \(x \to -3^-\). As \(x \to -3^+\): \(x > -3\) but less than -1, say \(x = -2.9\): \(x+3 = 0.1 > 0\), \(x-2 = -4.9 < 0\) → product negative. Numerator positive → \(f(x) \to -\infty\). So: \[ x \to -3^-: f(x) \to +\infty \] \[ x \to -3^+: f(x) \to -\infty \] --- At \(x = 2\): As \(x \to 2^-\): \(x < 2\) → \(x-2 < 0\), \(x+3 > 0\) → denominator negative. Numerator positive (since \(x=1.9\) gives \((0.9)(2.9)>0\)). So \(f(x) \to -\infty\). As \(x \to 2^+\): \(x > 2\) → \(x-2 > 0\), \(x+3 > 0\) → denominator positive, numerator positive → \(f(x) \to +\infty\). So: \[ x \to 2^-: f(x) \to -\infty \] \[ x \to 2^+: f(x) \to +\infty \] --- ### **Step 10: Behavior as \(x \to \pm\infty\)** Horizontal asymptote \(y = 1\). We can check if \(f(x) > 1\) or \(< 1\) for large \(|x|\): \[ f(x) - 1 = \frac{x^2 - 1}{x^2 + x - 6} - 1 = \frac{x^2 - 1 - (x^2 + x - 6)}{x^2 + x - 6} \] \[ = \frac{-x + 5}{x^2 + x - 6} \] For large \(x\), sign is like \(\frac{-x}{x^2} = -\frac{1}{x}\): - As \(x \to +\infty\), \(f(x) - 1 \to 0^-\), so \(f(x) \to 1\) from below. - As \(x \to -\infty\), \(f(x) - 1 \to 0^+\), so \(f(x) \to 1\) from above. --- ### **Step 11: Sketch** We have: 1. Vertical asymptotes: \(x = -3\) and \(x = 2\). 2. Horizontal asymptote: \(y = 1\). 3. x-intercepts: \((-1, 0)\), \((1, 0)\). 4. y-intercept: \((0, 1/6)\). 5. Sign intervals as above. From left: - As \(x \to -\infty\), curve approaches \(y=1\) from above. - At \(x=-3^-\), goes to \(+\infty\). - At \(x=-3^+\), from \(-\infty\). - Crosses x-axis at \(x=-1\) (going from negative to positive). - At \(x=1\), crosses x-axis again (going from positive to negative). - At \(x=2^-\), goes to \(-\infty\). - At \(x=2^+\), from \(+\infty\), then approaches \(y=1\) from below as \(x \to \infty\). --- **Final answer for the sketch description** (but since it’s text, I’ll summarize features): \[ \boxed{\text{VA: } x=-3, x=2; \quad \text{HA: } y=1; \quad x\text{-ints: } (-1,0),(1,0); \quad y\text{-int: } (0,1/6)} \]