1. Problem: Analyze the function $y = \log_2(x)$. - Domain: $x > 0$ because the logarithm is only defined for positive numbers. - Range: All real numbers $(-\infty, \infty)$. - Vertical asymptote: $x = 0$ since the function approaches $-\infty$ as $x$ approaches 0 from the right. - Horizontal asymptote: None. - Intercepts: Passes through $(1, 0)$ because $\log_2(1) = 0$. 2. Problem: Analyze the function $y = \log_3(x - 2)$. - Domain: $x > 2$ because the argument of the logarithm must be positive. - Range: All real numbers $(-\infty, \infty)$. - Vertical asymptote: $x = 2$. - Horizontal asymptote: None. - Intercepts: Passes through $(3, 0)$ since $\log_3(3 - 2) = \log_3(1) = 0$. 3. Problem: Analyze the function $y = 2^{x + 1} - 3$. - Domain: All real numbers $(-\infty, \infty)$. - Range: $(-3, \infty)$ because the exponential function is always positive and shifted down by 3. - Horizontal asymptote: $y = -3$. - Vertical asymptote: None. - Intercept: At $x=0$, $y = 2^{1} - 3 = 2 - 3 = -1$, so $(0, -1)$. 4. Problem: Analyze the function $y = \left(\frac{1}{5}\right)^x + 4$. - Domain: All real numbers $(-\infty, \infty)$. - Range: $(4, \infty)$ because the exponential decay approaches 0 and is shifted up by 4. - Horizontal asymptote: $y = 4$. - Vertical asymptote: None. - Intercept: At $x=0$, $y = 1 + 4 = 5$, so $(0, 5)$. 5. Problem: Analyze the function $y = -x^2 + 6x - 8$. - Domain: All real numbers $(-\infty, \infty)$. - Range: Since it opens downward, maximum at vertex. - Vertex: Use $x = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3$. - Evaluate $y$ at vertex: $y = -(3)^2 + 6(3) - 8 = -9 + 18 - 8 = 1$. - Range: $(-\infty, 1]$. - Intercepts: - $y$-intercept at $x=0$: $y = -0 + 0 - 8 = -8$, so $(0, -8)$. - $x$-intercepts: Solve $-x^2 + 6x - 8 = 0$ or $x^2 - 6x + 8 = 0$. $$ \begin{aligned} &x^2 - 6x + 8 = 0 \\ &\Rightarrow (x - 2)(x - 4) = 0 \\ &\Rightarrow x = 2, 4 \end{aligned} $$ So $x$-intercepts are $(2, 0)$ and $(4, 0)$. --- Summary of key points for plotting: (i) $y = \log_2(x)$: vertical asymptote $x=0$, passes $(1,0)$, $(2,1)$. (ii) $y = \log_3(x-2)$: vertical asymptote $x=2$, passes $(3,0)$, $(5,1)$. (iii) $y = 2^{x+1} - 3$: horizontal asymptote $y=-3$, passes $(0,-1)$. (iv) $y = (1/5)^x + 4$: horizontal asymptote $y=4$, passes $(0,5)$. (v) $y = -x^2 + 6x - 8$: vertex $(3,1)$, $x$-intercepts $(2,0)$ and $(4,0)$, $y$-intercept $(0,-8)$.