How do you graph #y<x-1 # and #y<=2x+1#?
Plot each graph separately and then shade the area that satisfies both inequalities.
Your equations are Step 1. Plot Use a solid line because your condition is "less than", not “less than or equal to".
Step 2. Plot Use a dashed line because your condition is "less than or equal to".
Step 3. Shade the area that satisfies both conditions. Shade the area that is below
Check Test a point in the coloured region. Try ( The point (
∴ TRUE
∴ TRUE
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To graph ( y < x - 1 ) and ( y \leq 2x + 1 ), follow these steps:
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Begin by graphing the line ( y = x - 1 ). This line has a slope of 1 (since the coefficient of x is 1) and a y-intercept of -1.
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Next, draw the line ( y = 2x + 1 ). This line also has a slope of 2 and a y-intercept of 1.
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Since ( y < x - 1 ), the region below the line ( y = x - 1 ) (not including the line itself) is shaded.
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Similarly, since ( y \leq 2x + 1 ), the region below or on the line ( y = 2x + 1 ) is shaded.
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The solution to the system of inequalities is the shaded region that satisfies both conditions.
Your graph should show two lines intersecting and the shaded area below both lines, including the region between them.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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