1. **Problem:** Sketch the graphs of the functions: a) $y = 2x - 9$ b) $y = x^2 - 4x + 3$ c) $y = -x^2 + 2x + 3$ **Step 1:** Understand the type of each function. - a) is a linear function with slope 2 and y-intercept -9. - b) is a quadratic function opening upwards (since coefficient of $x^2$ is positive). - c) is a quadratic function opening downwards (since coefficient of $x^2$ is negative). **Step 2:** Find key points for each graph. a) For $y=2x-9$: - y-intercept: $(0, -9)$ - x-intercept: solve $0=2x-9 \Rightarrow x=\frac{9}{2}=4.5$ b) For $y=x^2 -4x +3$: - Find vertex using $x=-\frac{b}{2a} = -\frac{-4}{2\times1} = 2$ - Vertex $y$ value: $2^2 -4\times2 +3 = 4 -8 +3 = -1$ - So vertex at $(2, -1)$ - Find roots by factoring: $x^2 -4x +3 = (x-3)(x-1)$, roots at $x=1$ and $x=3$ c) For $y=-x^2 + 2x + 3$: - Vertex $x = -\frac{b}{2a} = -\frac{2}{2\times(-1)} = 1$ - Vertex $y$ value: $-(1)^2 + 2\times1 + 3 = -1 + 2 + 3 = 4$ - Roots: solve $0 = -x^2 + 2x + 3 \Rightarrow x^2 - 2x - 3 = 0$ - Factor: $(x-3)(x+1) = 0$, roots at $x=3$ and $x=-1$ --- 2. **Problem:** Given piecewise function $$f(x) = \begin{cases} x+2 & x \leq 1 \\ x^2 + 1 & 1 < x \leq 2 \\ -x + 3 & x > 2 \end{cases}$$ Find the limits: a) $\lim_{x \to 1} f(x)$ b) $\lim_{x \to 2} f(x)$ c) $\lim_{x \to -1} f(x)$ **Step 1:** Calculate one-sided limits where needed. a) At $x=1$: - Left limit: $\lim_{x \to 1^-} f(x) = 1 + 2 = 3$ - Right limit: $\lim_{x \to 1^+} f(x) = 1^2 + 1 = 2$ - Since left and right limits differ, $\lim_{x \to 1} f(x)$ does not exist. b) At $x=2$: - Left limit: $\lim_{x \to 2^-} f(x) = 2^2 + 1 = 5$ - Right limit: $\lim_{x \to 2^+} f(x) = -2 + 3 = 1$ - Limits differ, so $\lim_{x \to 2} f(x)$ does not exist. c) At $x=-1$: - Since $-1 \leq 1$, use $f(x) = x + 2$ - So $\lim_{x \to -1} f(x) = -1 + 2 = 1$ --- 3. **Problem:** Calculate the following limits at infinity: a) $\lim_{x \to \infty} \frac{\sqrt{x^2 - x + 12}}{2x - 4}$ b) $\lim_{x \to \infty} \frac{-2x^2}{2x^2 + 3x}$ **Step 1:** Simplify each expression. a) For large $x$, inside the square root: $$\sqrt{x^2 - x + 12} = \sqrt{x^2\left(1 - \frac{1}{x} + \frac{12}{x^2}\right)} = |x| \sqrt{1 - \frac{1}{x} + \frac{12}{x^2}}$$ Since $x \to \infty$, $|x| = x$. So expression becomes: $$\frac{x \sqrt{1 - \frac{1}{x} + \frac{12}{x^2}}}{2x - 4} = \frac{x \sqrt{1 - \frac{1}{x} + \frac{12}{x^2}}}{x(2 - \frac{4}{x})}$$ Cancel $x$: $$\frac{\cancel{x} \sqrt{1 - \frac{1}{x} + \frac{12}{x^2}}}{\cancel{x}(2 - \frac{4}{x})} = \frac{\sqrt{1 - \frac{1}{x} + \frac{12}{x^2}}}{2 - \frac{4}{x}}$$ As $x \to \infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ go to 0: $$\lim_{x \to \infty} \frac{\sqrt{1 - 0 + 0}}{2 - 0} = \frac{1}{2}$$ b) Simplify: $$\frac{-2x^2}{2x^2 + 3x} = \frac{-2x^2}{x^2(2 + \frac{3}{x})} = \frac{-2 \cancel{x^2}}{\cancel{x^2}(2 + \frac{3}{x})} = \frac{-2}{2 + \frac{3}{x}}$$ As $x \to \infty$, $\frac{3}{x} \to 0$: $$\lim_{x \to \infty} \frac{-2}{2 + 0} = -1$$ --- **Final answers:** 1. Graphs: linear line $y=2x-9$, upward parabola $y=x^2 -4x +3$, downward parabola $y=-x^2 + 2x + 3$. 2. Limits of piecewise function: - a) $\lim_{x \to 1} f(x)$ does not exist. - b) $\lim_{x \to 2} f(x)$ does not exist. - c) $\lim_{x \to -1} f(x) = 1$. 3. Limits at infinity: - a) $\frac{1}{2}$ - b) $-1$