Question

ii. graphing and functions
a. linear graphs:

1. passes through the point (2, -1) and has the slope $-\frac{1}{3}$
2. passes through the point (4, -3) and is perpendicular to $3x + 2y = 4$
3. passes through the point (-1, -2) and is parallel to $y = \frac{3}{5}x - 1$

b. functions: find the domain of the following.

1. $f(x) = \frac{3}{x - 2}$ 24) $g(x) = \log(x - 3)$
2. $h(x) = \sqrt{2x - 3}$ 26) $w(x) = \frac{\sqrt{x - 1}}{x^2 - 1}$
3. given f(x) below, sketch the graph over the domain $-3, 3$.

$s(x) = \

$$\begin{cases} x & \\text{if } x \\geq 0 \\\\ 1 & \\text{if } -1 \\leq x < 0 \\\\ x - 2 & \\text{if } x < -1 \\end{cases}$$

$
find the composition/inverses as indicated below.
let $f(x) = x^2 + 3x - 2$ $g(x) = 4x - 3$ $h(x) = \ln x$ $w(x) = \sqrt{x - 4}$

1. $g^{-1}(x)$ 29) $h^{-1}(x)$ 30) $w^{-1}(x)$, for $x \geq 4$ 31) $f(g(x))$ 32) $h\left(g(f(1))\

ight)$

1. does $y = 3x^2 - 9$ have an inverse function? explain your answer.

let $f(x) = 2x$ $g(x) = -x$ $h(x) = 4$

1. $(f \circ g)(x)$ 35) $(f \circ g \circ h)(x)$
2. let $s(x) = \sqrt{4 - x}$ and $t(x) = x^2$, find the domain and range of $(s \circ t)(x)$.

c. basic shapes of curves:
sketch the graphs. you may use your graphing calculator to verify your graph, but you should be able to graph the following by knowledge of the shape of the curve, by plotting a few points, and by your knowledge of transformations.

1. $y = \sqrt{x}$ 38) $y = \ln x$ 39) $y = \frac{1}{x}$ 40) $y = |x - 2|$ 41) $y = \frac{1}{x - 2}$ 42) $\frac{x}{x^2 - 4}$
2. $y = e^{-x}$ 44) $f(x) = \

$$\begin{cases} \\sqrt{25 - x^2} & \\text{if } x < 0 \\\\ \\frac{x^2 - 25}{x - 5} & \\text{if } x \\geq 0, x \ eq 5 \\\\ 0 & \\text{if } x = 5 \\end{cases}$$

$

Explanation:

Step1: 确定使用的公式

已知直线过点\((x_0,y_0)\)且斜率为\(k\)，则直线的点斜式方程为\(y - y_0 = k(x - x_0)\)。本题中，点\((x_0,y_0)=(2,-1)\)，斜率\(k = -\frac{1}{3}\)。

Step2: 代入点斜式公式

将\(x_0 = 2\)，\(y_0 = -1\)，\(k = -\frac{1}{3}\)代入点斜式方程\(y - y_0 = k(x - x_0)\)，得到\(y - (-1)=-\frac{1}{3}(x - 2)\)。

Step3: 化简方程

对\(y + 1 = -\frac{1}{3}x+\frac{2}{3}\)进行化简，移项可得\(y = -\frac{1}{3}x+\frac{2}{3}-1\)，即\(y = -\frac{1}{3}x-\frac{1}{3}\)，也可整理为\(x + 3y + 1 = 0\)（两边同乘\(3\)：\(3y+3=-x + 2\)，移项得\(x + 3y+1 = 0\)）。

Answer:

直线方程为\(y = -\frac{1}{3}x-\frac{1}{3}\)（或\(x + 3y + 1 = 0\)）