Question

parts of linear equations(a1.3a)

1. find the slope and y-intercept. then graph.

$y = \frac{1}{2}x - 3$
$m=\\_\\_\\_\\_\\_$
$b=\\_\\_\\_\\_\\_$

1. find the slope and y-intercept. then graph.

$y = x - 1$
$m=\\_\\_\\_\\_\\_$
$b=\\_\\_\\_\\_\\_$

1. find the slope and y-intercept. then graph.

$y = \frac{1}{3}x + 1$
$m=\\_\\_\\_\\_\\_$
$b=\\_\\_\\_\\_\\_$

1. find the slope and y-intercept. then graph.

$y = x$
$m=\\_\\_\\_\\_\\_$
$b=\\_\\_\\_\\_\\_$
changes in y=mx+b (a1.3e)

1. which of the following is 2 units higher than $y=3x - 1$? (circle one)

a) $y = 3x + 1$
b. $y = 5x - 1$

1. which of the following equations has the same slope as $y = \frac{1}{3}x - 3$? (circle one)

a) $y = \frac{1}{3}x - 3$
b. $y = 3x + 5$

1. which slope is the steepest? (circle one)

a. $y = 4x - 3$
b. $y = 1/3x - 4$

Explanation:

Problem 10:

Step1: Recall slope - intercept form

The slope - intercept form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept.

Step2: Identify $m$ and $b$ for $y=x - 1$

For the equation $y=x - 1$, we can rewrite it as $y = 1x+( - 1)$. Comparing with $y=mx + b$, we have $m = 1$ and $b=-1$.

Step1: Recall slope - intercept form

The slope - intercept form of a linear equation is $y=mx + b$, where $m$ is the slope and $b$ is the y - intercept.

Step2: Identify $m$ and $b$ for $y = x$

For the equation $y=x$, we can rewrite it as $y = 1x+0$. Comparing with $y = mx + b$, we have $m = 1$ and $b = 0$.

Step1: Analyze the change in the equation

To find a line that is 2 units higher than $y = 3x-1$, we need to change the y - intercept. The general form is $y=mx + b$. For a vertical shift of $k$ units up, the new equation is $y=mx+(b + k)$. Here, $m = 3$, $b=-1$ and $k = 2$.

Step2: Calculate the new $b$

The new $b$ value is $b_{new}=-1 + 2=1$. So the equation of the line that is 2 units higher than $y = 3x - 1$ is $y=3x + 1$. So the correct option is a. $y = 3x+1$

Answer:

$m = 1$, $b=-1$

Problem 12: