nico is trying to write an equation in slope-intercept form for a line through the point (1,2) and has a slope of -1/2 what is the y-intercept of this line rounded to the nearest tenth?

To find the equation of a line in slope-intercept form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept, we can use the slope (-1/2) and the point (1, 2).

We have:

• \( m = -\frac{1}{2} \)
• The point \( (x_1, y_1) = (1, 2) \)

First, we can substitute the slope and the coordinates of the point into the equation \( y = mx + b \):

1. Start with \( y = mx + b \): \[ y = -\frac{1}{2}x + b \]

2. Substitute the coordinates of the point \( (1, 2) \): \[ 2 = -\frac{1}{2}(1) + b \] \[ 2 = -\frac{1}{2} + b \]

3. To isolate \( b \), add \( \frac{1}{2} \) to both sides: \[ 2 + \frac{1}{2} = b \] \[ b = \frac{4}{2} + \frac{1}{2} = \frac{5}{2} \]

Now convert \( \frac{5}{2} \) to a decimal: \[ \frac{5}{2} = 2.5 \]

Thus, the y-intercept \( b \) is \( 2.5 \).

So the y-intercept of this line, rounded to the nearest tenth, is \( \boxed{2.5} \).