### 数学代写|微积分代写Calculus代写|Slope-intercept form of the equation of a line

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• Foundations of Data Science 数据科学基础

## 数学代写|微积分代写Calculus代写|Slope-intercept form of the equation of a line

Suppose we know that a line has slope $m$ and $y$-intercept $b$. The $y$ intercept gives the value of $y$ at which the line crosses the $y$-axis, so the point $(0, b)$ is on the graph of the line (figure 20$)$.

Using the point-slope form of the equation of a line with slope $m$ and point $\left(x_{1}, y_{1}\right)=(0, b)$ gives
\begin{aligned} y-y_{1} &=m\left(x-x_{1}\right) \ y-b &=m(x-0) \ y-b &=m x \ y &=m x+b . \end{aligned}The traditional form of the answer as used in examples 5 and 6 is the slope-intercept form. That makes it easy to recognize the $y$-intercept;

for the equation $y=-2 x+6$, the $y$-intercept is 6 and the slope is $-2$. For the equation $y=-\frac{3}{4} x+\frac{11}{4}$, the $y$-intercept is $\frac{11}{4}$ and the slope is $-\frac{3}{4}$. This combination of information is all we need to graph the line relatively quickly.
Example 7 Graph the line $y=2 x-3$.
Solution Instead of plotting points as in example 2, we use the slope and $y$-intercept of the line to mark points quickly on the graph. The line is in slope-intercept form, with slope $m=2$ and $y$-intercept $-3$. We begin by placing the $y$-intercept on the graph (figure 21).
Next we interpret the slope as rise:
$$\text { slope }=\frac{\text { rise }}{\text { run }}=\frac{2}{1},$$
making the rise 2 when the run is 1 . Starting at the $y$-intercept, we move to the right 1 unit and up 2 units to find another point on the graph, and repeat as often as desired (figure 22).

## 数学代写|微积分代写Calculus代写|Parallel and perpendicular lines

Parallel lines have the same slope. Since $y=-4 x-29$ has slope $-4$, any line parallel to $y=-4 x-29$ has slope $-4$.

Example 8 Find the equation of the line through $(10,4)$ parallel to $y=$ $7 x+1$.

Solution Whichever form of the equation of a line is used, the point-slope form or the slope-intercept form, the slope is required information. Our line must be parallel to $y=7 x+1$, which has slope $m=7$; therefore, the slope of our line is $m=7$. We have a point and

the slope, so we can use the point-slope form of the equation of a line:
\begin{aligned} y-y_{1} &=m\left(x-x_{1}\right) \ y-4 &=7(x-10) \ y-4 &=7 x-70 \ y &=7 x-66 \end{aligned}
The equation of the line is $y=7 x-66$.
Perpendicular lines have slopes that are negative reciprocals. If the slopes of the lines are $m_{1}$ and $m_{2}$, then $m_{2}=-\frac{1}{m_{1}}$. See figure 25 .

## 数学代写|微积分代写Calculus代写|Vertical lines

Vertical lines consist of points with the same $x$-coordinate, so their equations have the form $x=c$ for a real number $c$ (see figure 16).
Example 10 Find the equation of the vertical line through the point $(5,3)$.

Solution All points on a vertical line have the same $x$-coordinate. The given point on the line has $x$-coordinate 5 . Therefore, all points on the line have $x$-coordinate 5 . The equation of the line is $x=5$.
Exercises $\mathbf{0 . 2}$
1-12. Rapid response: state the slope and $y$-intercept of the line.

1. $y=4 x-7$
2. $y+1=5 x$
3. $y=2 x+5$
4. $y=\frac{1}{4} x-15$
5. $y=-91 x+225$
6. $y=\frac{2 x}{3}+12$
7. $y=-3 x-11$
8. $y-3=-2 x$
9. $y=17 x$
10. $y=2(x+3)$
11. $y=17$
12. $y=-(x+4)$
13. Plot the points $(4,7),\left(-1, \frac{3}{2}\right)$, and $(0,5)$.
14. Plot the points $(-2,0),(6,-1)$, and $\left(2, \frac{1}{2}\right)$.
15-24. Graph the equation.
15. $y=x^{2}$
16. $y=\frac{x^{2}}{3}$
17. $y=x^{2}-5$
18. $y=3 x^{2}-11$
19. $y=7-x^{2}$
20. $y=\frac{x+1}{2}$
21. $y=\sqrt{x}$
22. $y=1+\sqrt{x-2}$
23. $y=\sqrt{x+2}$
24. $y=x^{3}$

坡 = 上升  跑 =21,

## 数学代写|微积分代写Calculus代写|Vertical lines

1-12。快速响应：说明斜率和是-截取线。

1. 是=4X−7
2. 是+1=5X
3. 是=2X+5
4. 是=14X−15
5. 是=−91X+225
6. 是=2X3+12
7. 是=−3X−11
8. 是−3=−2X
9. 是=17X
10. 是=2(X+3)
11. 是=17
12. 是=−(X+4)
13. 绘制点(4,7),(−1,32)， 和(0,5).
14. 绘制点(−2,0),(6,−1)， 和(2,12).
15-24。绘制方程。
15. 是=X2
16. 是=X23
17. 是=X2−5
18. 是=3X2−11
19. 是=7−X2
20. 是=X+12
21. 是=X
22. 是=1+X−2
23. 是=X+2
24. 是=X3

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## MATLAB代写

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