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How do you graph #y < 7/3x + 52/3#?

Answer 1

Draw a dotted straight line graph and shade the area below the graph.

Graphing a linear inequality is the same as graphing a straight line.

In this case, the #y#-intercept is #17 1/3# and the slope is #7/3#

The straight line will be drawn as a dotted line. The shaded area indicating the required region will be BELOW the line.

Use the origin as a point to check:

For #(0,0)#, in the equation #" "y < 1/3x+52/3#

#0 < 51 1/3" "larr# this is true, so the origin lies in the required region.

graph{y< 7/3x+52/3 [-28.54, 51.46, -7.88, 32.12]}

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Answer 2

Genesis Allen

To graph the inequality (y < \frac{7}{3}x + \frac{52}{3}), follow these steps:

Start by graphing the boundary line (y = \frac{7}{3}x + \frac{52}{3}) as if it were an equation.
Since the inequality is (y < \frac{7}{3}x + \frac{52}{3}), the solution includes all points below the boundary line.
Use a dashed line to graph the boundary line since the inequality does not include points on the line itself.
Shade the area below the dashed line to represent the solution set.

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Answer from HIX Tutor

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.

How do you graph #y &lt; 7/3x + 52/3#?

How do you graph #y < 7/3x + 52/3#?