Question

name:_________________________
test 3: unit 3 linear functions
eq1: what is the significance of slope and y - intercept in linear functions?
eq2: how can we represent real - world situations with linear functions?
directions: show all work, when possible, to receive full credit.
slope formula
$m = \frac{y_2 - y_1}{x_2 - x_1}$
slope - intercept form
$y = mx + b$
point - slope formula
$y - y_1 = m(x - x_1)$

1. find the slope of the line that passes through the following points.

a. $(3, -5)$ and $(9, -3)$
$(4, 3)$ and $(-7, 3)$

1. graph the following linear functions.

a. $y = \frac{-2}{3}x + 5$
b. $-2y + 3x = 2$

1. graph the linear inequalities, state one solution to the inequality.

a. $y > 3x + 4$
b. $3x - y \geq 3$

Explanation:

Problem 1a: Find the slope of the line through (3, -5) and (9, -3)

Step1: Identify coordinates

Let \((x_1, y_1) = (3, -5)\) and \((x_2, y_2) = (9, -3)\).

Step2: Apply slope formula

Slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - (-5)}{9 - 3}\)

Step3: Simplify numerator and denominator

\(m = \frac{-3 + 5}{6} = \frac{2}{6} = \frac{1}{3}\)

Step1: Identify coordinates

Let \((x_1, y_1) = (4, 3)\) and \((x_2, y_2) = (-7, 3)\).

Step2: Apply slope formula

Slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 3}{-7 - 4}\)

Step3: Simplify numerator and denominator

\(m = \frac{0}{-11} = 0\)

Step1: Identify slope and y-intercept

Slope \(m = -\frac{2}{3}\), y-intercept \(b = 5\) (so the line crosses the y-axis at (0, 5)).

Step2: Plot y-intercept

Mark the point (0, 5) on the y-axis.

Step3: Use slope to find next point

From (0, 5), move down 2 units (since slope is \(-\frac{2}{3}\), rise = -2) and right 3 units (run = 3) to get (3, 3).

Step4: Draw the line

Connect (0, 5) and (3, 3) with a straight line (extending both ways).

Answer:

\(\frac{1}{3}\)

Problem 1b: Find the slope of the line through (4, 3) and (-7, 3)