1. **State the problem:** Determine the domain, range, and whether each given graph represents a function. - **Top-left graph:** Vertical asymptote near $x=2$ and horizontal asymptote near $y=0$, branches in first and third quadrants (rational function). - **Top-right graph:** Sinusoidal wave with multiple x-axis crossings (periodic sine or cosine function). - **Bottom-left graph:** Piecewise linear "V" shape rising from bottom-left to top-right. - **Bottom-right:** No graph. **Step 1:** Identify domain for each graph: - Top-left: All real numbers except where vertical asymptote occurs, so $\text{Domain} = (-\infty, 2) \cup (2, \infty)$. - Top-right: Sinusoidal wave continuous for all real $x$, so $\text{Domain} = (-\infty, \infty)$. - Bottom-left: Piecewise linear with no breaks, so $\text{Domain} = (-\infty, \infty)$. **Step 2:** Identify range: - Top-left: Horizontal asymptote $y=0$ approached but not crossed, branches in first and third quadrants mean range is $(-\infty, 0) \cup (0, \infty)$. - Top-right: Sinusoidal wave oscillates between max and min, typically $[-1, 1]$ or similar; given no exact numbers, assume $\text{Range} = [-1,1]$. - Bottom-left: V shape opens upward from bottom-left to top-right, minimum at vertex, so range starts at minimum value. Assuming vertex at origin: $\text{Range} = [0, \infty)$. **Step 3:** Determine if each relation is a function: - Top-left: Rational function with vertical asymptote, passes vertical line test, yes, function. - Top-right: Sinusoidal wave, continuous function, yes. - Bottom-left: V shape, piecewise linear, also function. **Answers for item 1:** - Top-left: Domain $(-\infty, 2) \cup (2, \infty)$; Range $(-\infty, 0) \cup (0, \infty)$; Function? Yes. - Top-right: Domain $(-\infty, \infty)$; Range $[-1,1]$; Function? Yes. - Bottom-left: Domain $(-\infty, \infty)$; Range $[0, \infty)$; Function? Yes. 2. **Find the value of $f(-7)$:** No explicit function $f$ given in item 2, so unable to evaluate $f(-7)$. Possibly use in item 3. 3. **Evaluate $f(2x + 3)$ for $f(x) = -x^2 + 3x + 10$:** Step 1: Substitute $x$ with $2x+3$ in the function: $$f(2x+3) = -(2x+3)^2 + 3(2x+3) + 10$$ Step 2: Expand $(2x+3)^2$: $$(2x+3)^2 = 4x^2 + 12x + 9$$ Step 3: Substitute and simplify: $$f(2x+3) = -(4x^2 + 12x + 9) + 6x + 9 + 10$$ $$= -4x^2 - 12x - 9 + 6x + 9 + 10$$ Step 4: Combine like terms: $$-4x^2 - 12x + 6x -9 + 9 + 10 = -4x^2 - 6x + 10$$ **Final answer for item 3:** $$f(2x+3) = -4x^2 -6x + 10$$ 4. **Find zero(s) and y-intercept for $f(x) = 4x^4 - 17x^2 + 4$:** Step 1: Find zeros by solving $4x^4 - 17x^2 + 4 = 0$. Step 2: Use substitution let $u = x^2$, then quadratic in $u$: $$4u^2 - 17u + 4 = 0$$ Step 3: Solve quadratic using quadratic formula: $$u = \frac{17 \pm \sqrt{(-17)^2 - 4 \times 4 \times 4}}{2 \times 4} = \frac{17 \pm \sqrt{289 - 64}}{8} = \frac{17 \pm \sqrt{225}}{8}$$ $$= \frac{17 \pm 15}{8}$$ Step 4: Calculate roots: $$u_1 = \frac{17+15}{8} = \frac{32}{8} = 4$$ $$u_2 = \frac{17-15}{8} = \frac{2}{8} = \frac{1}{4}$$ Step 5: Back-substitute $u=x^2$: $$x^2 = 4 \Rightarrow x = \pm 2$$ $$x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$$ Step 6: Y-intercept is $f(0)$: $$f(0) = 4(0)^4 - 17(0)^2 + 4 = 4$$ **Answer for item 4:** Zero(s): $x = -2, -\frac{1}{2}, \frac{1}{2}, 2$; Y-intercept: $4$ 5. **Find zero(s) and y-intercept for $f(x) = x^6 - 81x^4$:** Step 1: Factor the function: $$f(x) = x^4(x^2 - 81) = x^4(x - 9)(x + 9)$$ Step 2: Solve for zeros: $$x^4 = 0 \Rightarrow x=0$$ $$x-9=0 \Rightarrow x=9$$ $$x+9=0 \Rightarrow x=-9$$ Step 3: Y-intercept is $f(0)$: $$f(0) = 0 - 0 = 0$$ **Answer for item 5:** Zero(s): $x = -9, 0, 9$; Y-intercept: $0$