Question

1. write an equation in point - slope form where ( m=-4 ) and goes through the point ( (3,-2) ).
2. circle the four equations that represent the line.

( f(x)=-\frac{2}{3}x + 2 )
( f(x)=-\frac{3}{2}x + 2 )
( y=-\frac{2}{3}x + 2 )
( y + 2=-\frac{2}{3}(x - 6) )
( y + 2=-\frac{3}{2}(x - 6) )
( y - 4=-\frac{2}{3}(x + 3) )

1. write the equation in slope - intercept form:

( y + 7=3(x - 2) )

Explanation:

Step1: Find the slope

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1,y_1)=(-3,4)\) and \((x_2,y_2)=(6,-2)\).
\(m=\frac{-2 - 4}{6-(-3)}=\frac{-6}{9}=-\frac{2}{3}\)

Step2: Check each equation

• For \(a\): \(f(x)=-\frac{2}{3}x + 2\), slope \(m =-\frac{2}{3}\). Substitute \(x = 6\), \(f(6)=-\frac{2}{3}\times6+2=-4 + 2=-2\). Substitute \(x=-3\), \(f(-3)=-\frac{2}{3}\times(-3)+2=2 + 2=4\).
• For \(b\): \(f(x)=-\frac{3}{2}x + 2\), slope \(m=-\frac{3}{2}

eq-\frac{2}{3}\).

• For \(c\): \(y=-\frac{2}{3}x + 2\), same as \(a\) (just different notation \(y = f(x)\)).
• For \(d\): \(y + 2=-\frac{2}{3}(x - 6)\), expand \(y=-\frac{2}{3}x+4 - 2=-\frac{2}{3}x+2\).
• For \(e\): \(y + 2=-\frac{3}{2}(x - 6)\), slope \(m =-\frac{3}{2}

eq-\frac{2}{3}\).

• For \(f\): \(y - 4=-\frac{2}{3}(x + 3)\), expand \(y=-\frac{2}{3}x-2 + 4=-\frac{2}{3}x+2\).

Answer:

\(a\), \(c\), \(d\), \(f\)