91Ó°ÊÓ

The given linear function is \( h(t) = \frac{1}{2} - \frac{3}{4}t \). This function is in the form of \( y = mx + b \) where \( m \) is the slope and \( b \) is the y-intercept. For this function, the y-intercept \( b \) is \( \frac{1}{2} \), and the slope \( m \) is \(-\frac{3}{4}\).

To graph the function, we'll create a table of values. Choose several values for \( t \) (for example, \(-2, -1, 0, 1, 2\)) and calculate \( h(t) \) for each. - \( t = -2 \), \( h(-2) = \frac{1}{2} - \frac{3}{4}(-2) = \frac{1}{2} + \frac{3}{2} = 2 \)- \( t = -1 \), \( h(-1) = \frac{1}{2} - \frac{3}{4}(-1) = \frac{1}{2} + \frac{3}{4} = \frac{5}{4} \)- \( t = 0 \), \( h(0) = \frac{1}{2} - \frac{3}{4}(0) = \frac{1}{2} \)- \( t = 1 \), \( h(1) = \frac{1}{2} - \frac{3}{4}(1) = \frac{1}{2} - \frac{3}{4} = -\frac{1}{4} \)- \( t = 2 \), \( h(2) = \frac{1}{2} - \frac{3}{4}(2) = \frac{1}{2} - \frac{3}{2} = -1 \)The table of values is:| \( t \) | \( h(t) \) ||-------|-------|| -2 | 2 || -1 | \( \frac{5}{4} \) || 0 | \( \frac{1}{2} \) || 1 | \( -\frac{1}{4} \) || 2 | -1 |

Using the table of values, plot the points \((-2, 2)\), \((-1, \frac{5}{4})\), \((0, \frac{1}{2})\), \((1, -\frac{1}{4})\), and \((2, -1)\) on a coordinate plane. Draw a straight line through these points, extending it in both directions. This line represents the graph of the linear function \( h(t) = \frac{1}{2} - \frac{3}{4}t \).

The slope of a linear function in the form \( y = mx + b \) is represented by \( m \). For the function \( h(t) = \frac{1}{2} - \frac{3}{4}t \), the slope is \(-\frac{3}{4}\). This indicates that for every unit increase in \( t \), \( h(t) \) decreases by \( \frac{3}{4} \).