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Question
extra example 1: write an equation for the line of fit drawn.
Step1: Identify two points on the line
From the graph, the line passes through (1, 30) and (10, 2) (assuming the last point is (10, 2) as it's near y=2 at x=10).
Step2: Calculate the slope (m)
Using the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), substitute \( (x_1, y_1) = (1, 30) \) and \( (x_2, y_2) = (10, 2) \).
\( m = \frac{2 - 30}{10 - 1} = \frac{-28}{9} \approx -3.11 \)? Wait, maybe better points. Let's take (1, 30) and (10, 2) is wrong. Wait, looking at the line, when x=1, y=30; x=10, y=2? No, maybe (0, 30) and (10, 2)? Wait, x=1, y=30; x=10, y=2? Wait, maybe another pair. Let's see, when x=1, y=30; x=5, y=20? Wait, no, the line of fit: let's take two clear points. Let's say (1, 30) and (10, 2) is incorrect. Wait, maybe (1, 30) and (10, 2) is wrong. Wait, looking at the graph, the line starts at (1, 30) and ends at (10, 2)? No, maybe (0, 30) and (10, 2) but x starts at 1. Wait, maybe (1, 30) and (10, 2) is not right. Wait, let's calculate slope between (1, 30) and (10, 2): \( m = (2 - 30)/(10 - 1) = -28/9 ≈ -3.11 \). But maybe better to take (1, 30) and (6, 15)? Wait, no. Wait, the line of fit: let's see, when x=1, y=30; x=10, y=2. So slope is (2 - 30)/(10 - 1) = -28/9 ≈ -3.11. But maybe the line is y = -3x + 33? Wait, let's check x=1: y=-3(1)+33=30, correct. x=10: y=-3(10)+33=3, close to the last point. Maybe the slope is -3. Let's recheck. If x=1, y=30; x=2, y=27; x=3, y=24; so slope is -3. So slope m = -3. Then using point-slope form: y - y1 = m(x - x1). Using (1, 30): y - 30 = -3(x - 1) → y = -3x + 3 + 30 → y = -3x + 33. Let's check x=10: y = -30 + 33 = 3, which is close to the last point. So that works.
Step3: Write the equation
Using slope-intercept form y = mx + b. We found m = -3, and when x=1, y=30: 30 = -3(1) + b → b = 33. So equation is y = -3x + 33.
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The equation of the line of fit is \( y = -3x + 33 \) (or other equivalent forms based on accurate point selection, but this is a reasonable approximation).