1. **Problem statement:** Given the function $f(x) = \frac{n(x)}{d(x)}$ where $f$ touches but does not cross the x-axis at $x = -2$, and the function $g(x) = \frac{ax^3 + bx^2 + 6}{x^2 + x + 1}$ with a slant asymptote $y = 2x + 5$, answer the following questions. 2. **(a) Common factor in $n(x)$ and $d(x)$:** Since $f$ touches but does not cross the x-axis at $x = -2$, $x = -2$ is a root of $n(x)$ with even multiplicity, and also a root of $d(x)$ (to prevent crossing). Therefore, the factor $(x + 2)$ is common to both $n(x)$ and $d(x)$. 3. **(b) Minimum degree of $d(x)$:** Since $f$ has vertical asymptotes near $x = -1$ and $x = 1$, $d(x)$ must have factors $(x + 1)$ and $(x - 1)$. Also, since $(x + 2)$ is a common factor with $n(x)$, $d(x)$ must have at least degree 3 to include these three factors. Thus, minimum degree of $d(x)$ is 3. 4. **(c) Horizontal asymptote of $j(x) = -2f(3(x-1)) + 4$:** The horizontal asymptote of $f(x)$ is $y = 3$ (given by the dashed line). Calculate horizontal asymptote of $j(x)$: $$\lim_{x \to \infty} j(x) = -2 \cdot \lim_{x \to \infty} f(3(x-1)) + 4 = -2 \cdot 3 + 4 = -6 + 4 = -2$$ So, the horizontal asymptote of $j$ is $y = -2$. 5. **(d) Find $a$ and $b$ given slant asymptote $y = 2x + 5$ for $g(x)$:** Perform polynomial division of numerator by denominator: Divide $ax^3 + bx^2 + 6$ by $x^2 + x + 1$: - Leading term division: $\frac{ax^3}{x^2} = ax$ - Multiply denominator by $ax$: $ax^3 + ax^2 + ax$ - Subtract: $(ax^3 + bx^2 + 6) - (ax^3 + ax^2 + ax) = (b - a)x^2 - ax + 6$ Next term: $\frac{(b - a)x^2}{x^2} = b - a$ Multiply denominator by $b - a$: $(b - a)x^2 + (b - a)x + (b - a)$ Subtract: $$((b - a)x^2 - ax + 6) - ((b - a)x^2 + (b - a)x + (b - a)) = (-a - (b - a))x + (6 - (b - a)) = (-a - b + a)x + (6 - b + a) = -b x + (6 - b + a)$$ Since slant asymptote is $y = 2x + 5$, the quotient must be $ax + (b - a) = 2x + 5$. Equate coefficients: $$a = 2$$ $$b - a = 5 \implies b - 2 = 5 \implies b = 7$$ 6. **(e) Vertical asymptotes of $g(x)$:** Vertical asymptotes occur where denominator is zero: $$x^2 + x + 1 = 0$$ Discriminant: $$\Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$$ No real roots, so no vertical asymptotes. **Final answers:** - (a) Common factor: $(x + 2)$ - (b) Minimum degree of $d(x)$: 3 - (c) Horizontal asymptote of $j$: $y = -2$ - (d) $a = 2$, $b = 7$ - (e) No vertical asymptotes