1. Find the domain and range of each relation using interval notation, and determine if it is a function. - Top-left graph (rational function with vertical asymptote at $x=1$ and horizontal asymptote at $y=0$): - Domain excludes $x=1$ where the vertical asymptote is, so: $$\text{Domain} = (-\infty, 1) \cup (1, \infty)$$ - Range excludes $y=0$ because of horizontal asymptote: $$\text{Range} = (-\infty, 0) \cup (0, \infty)$$ - Function? Yes, since each $x$ corresponds to exactly one $y$ except $x=1$ which is excluded. - Top-right graph (sinusoidal wave): - Domain covers all real numbers: $$\text{Domain} = (-\infty, \infty)$$ - Range oscillates between minimum and maximum values, typically between $-1$ and $1$, so: $$\text{Range} = [-1, 1]$$ - Function? Yes, the sinusoidal graph passes vertical line test. - Bottom-left graph (piecewise linear, downward then upward slope): - Domain covers all real numbers (graph extends left and right indefinitely): $$\text{Domain} = (-\infty, \infty)$$ - Range depends on the values the graph attains, assuming minimum near zero and extending upwards and downwards, plausible approximate range: $$\text{Range} = (-\infty, \infty)$$ - Function? Yes, each input has only one output. 2. Find $f(-7)$ for $f(x)$ defined by the top-left graph. Since the top-left graph is a rational function with vertical asymptote at $x=1$. The function likely behaves as $f(x) = \frac{1}{x-1}$ or similar. Evaluate $f(-7)$: $$f(-7) = \frac{1}{-7-1} = \frac{1}{-8} = -\frac{1}{8}$$ 3. Evaluate $f(2x + 3)$ where $f(x) = -x^2 + 3x + 10$. 1. Substitute $2x + 3$ in place of $x$: $$f(2x + 3) = - (2x + 3)^2 + 3(2x + 3) + 10$$ 2. Expand the square: $$(2x + 3)^2 = 4x^2 + 12x + 9$$ 3. Substitute back: $$= - (4x^2 + 12x + 9) + 6x + 9 + 10$$ 4. Distribute the negative: $$= -4x^2 -12x -9 + 6x + 9 + 10$$ 5. Combine like terms: $$= -4x^2 -6x + 10$$ 4. Find zeros and y-intercept for $f(x) = 4x^4 - 17x^2 + 4$. 1. Set $f(x) = 0$: $$4x^4 - 17x^2 + 4 = 0$$ 2. Substitute $u = x^2$: $$4u^2 - 17u + 4 = 0$$ 3. Solve quadratic using quadratic formula: $$u = \frac{17 \pm \sqrt{(-17)^2 - 4 \times 4 \times 4}}{2 \times 4} = \frac{17 \pm \sqrt{289 - 64}}{8} = \frac{17 \pm \sqrt{225}}{8} = \frac{17 \pm 15}{8}$$ 4. Calculate roots: $$u_1 = \frac{17 + 15}{8} = 4$$ $$u_2 = \frac{17 - 15}{8} = \frac{2}{8} = \frac{1}{4}$$ 5. Substitute back $x^2 = u$: $$x^2 = 4 \rightarrow x = \pm 2$$ $$x^2 = \frac{1}{4} \rightarrow x = \pm \frac{1}{2}$$ Zeros are $x = -2, -\frac{1}{2}, \frac{1}{2}, 2$. 6. Y-intercept is $f(0)$: $$f(0) = 4\times0 - 17\times0 + 4 = 4$$ 5. Find zeros and y-intercept for $f(x) = x^6 - 81x^4$. 1. Factor: $$f(x) = x^4(x^2 - 81) = x^4(x - 9)(x + 9)$$ 2. Set each factor to zero: $$x^4 = 0 \Rightarrow x = 0$$ $$x - 9 = 0 \Rightarrow x = 9$$ $$x + 9 = 0 \Rightarrow x = -9$$ Zeros: $x = -9, 0, 9$ 3. Y-intercept: $$f(0) = 0^6 - 81\times 0^4 = 0$$