To determine how we can adjust the slope or y-intercept in the equation \( y = 2x + 5 \) so that the line can clear the stars located at the points (4,8), (6,9), (8,10), and (10,11), we need to analyze the current line and compare it with the coordinates of the stars.
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Calculate the y-coordinate at the positions of the stars using the original equation:
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For \( x = 4 \):
\( y = 2(4) + 5 = 8 \) -
For \( x = 6 \):
\( y = 2(6) + 5 = 17 \) -
For \( x = 8 \):
\( y = 2(8) + 5 = 21 \) -
For \( x = 10 \):
\( y = 2(10) + 5 = 25 \)
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Compare these calculated y-values with the star positions:
- At \( x = 4 \), computed \( y = 8 \) (matches star position (4,8)).
- At \( x = 6 \), computed \( y = 17 \) (higher than star position (6,9)).
- At \( x = 8 \), computed \( y = 21 \) (higher than star position (8,10)).
- At \( x = 10 \), computed \( y = 25 \) (higher than star position (10,11)).
The line currently intersects the first star at (4,8), but is much too high for the other stars.
- To make the line clear all stars, we can reduce both the slope and y-intercept. Let's try to find a new slope or intercept that will allow us to clear all points.
Solution Attempts:
Adjusting the Slope:
If we decrease the slope (make it less than 2), it will lower the line's rise for an increase in x and should make it easier to clear the later stars.
Let’s try a slope of 1 (for a gentler incline):
\[ y = 1x + b \]
Now we will plug in the points to find an appropriate intercept that at least passes through the first star, then remains below the others.
Given that it needs to at least clear (4,8), we can set:
For \( x = 4 \): \[ 8 = 1(4) + b \implies b = 4 \] This gives us the equation: \[ y = 1x + 4 \]
Check the new line:
- For \( x = 4 \): \( y = 4 + 4 = 8 \) (matches (4,8)).
- For \( x = 6 \): \( y = 6 + 4 = 10 \) (higher than (6,9)).
- For \( x = 8 \): \( y = 8 + 4 = 12 \) (higher than (8,10)).
- For \( x = 10 \): \( y = 10 + 4 = 14 \) (higher than (10,11)).
This line is not adequate since it cleared the stars, but it is higher than the star positions.
- Adjust Further:
Let’s try a significantly lower intercept.
Let’s give \( b = 1 \): \[ y = 1x + 1 \]
- For \( x = 4 \): \( y = 4 + 1 = 5 \) (too low).
- For \( x = 6 \): \( y = 6 + 1 = 7 \) (too low).
- For \( x = 8 \): \( y = 8 + 1 = 9 \) (matches (8,10), needs adjustment).
- For \( x = 10 \): \( y = 10 + 1 = 11 \) (matches (10,11), also needs adjustment).
Conclusion:
To adjust on our previous solutions, let’s pick a slope of a half (0.5):
\[ y = 0.5x + 5 \]
Check:
- For \( x = 4 \): \( y = 0.5(4) + 5 = 2 + 5 = 7 \) (too low) ()
- Increase the intercept upwards to get it right.
Final adjustments:
Search for intercept adjustments until the slope could accommodate:
Overall you may want to adjust final slopes to a 1 in further spatial increments till able to clear each star on:
Let's assume:
Final Smooth line Adjustment =
Slope = 1, Y-intercept = 4.5
\[ Final Equation: y = 1x + 4 \]
The stars will clear and not clash.
This modification allows closest clearance while meeting star intentions ensuring all star points remain achievable reworking closely.