Question

function 1 is defined by the equation $y = \frac{1}{3}x + \frac{1}{2}$. function 2 is defined by line $q$, shown on the following graph. graph of line $q$ on a coordinate plane with x-axis from -8 to 8 and y-axis from -8 to 8, passing through the origin and other points which function has a greater slope?

Explanation:

Step1: Find slope of Function 1

Function 1 is \( y = \frac{1}{3}x + \frac{1}{2} \), in slope - intercept form \( y=mx + b \) (where \( m \) is slope), so slope \( m_1=\frac{1}{3}\approx0.333 \).

Step2: Find slope of Function 2

For Function 2, the line passes through \((0,0)\) and \((6,1)\) (we can also use other points like \((- 6,-1)\)). The slope formula is \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Using \((x_1,y_1)=(0,0)\) and \((x_2,y_2)=(6,1)\), we get \( m_2=\frac{1 - 0}{6 - 0}=\frac{1}{6}\approx0.167 \).

Step3: Compare the slopes

Compare \( \frac{1}{3} \) and \( \frac{1}{6} \). Since \( \frac{1}{3}=\frac{2}{6} \) and \( \frac{2}{6}>\frac{1}{6} \), so \( m_1>m_2 \).

Answer:

Function 1