Graph the following inequality:
expr([ "+", [ "*", A, "x" ], [ "*", B, "y" ] ]) STD_FORM_COMP C
y COMP PRETTY_SLOPE + YINT
Convert expr([ "+", [ "*", A, "x" ], [ "*", B, "y" ] ]) STD_FORM_COMP C
to slope-intercept form by solving for y
.
Add abs( A )x
to both sides:
Subtract abs( A )x
from both sides:
\qquad expr( [ "*", B, "y" ] ) STD_FORM_COMP expr([ "+", [ "*", -A, "x" ], C ])
Divide both sides by B
. Since you're multiplying or dividing by a negative number, don't forget to flip the inequality sign:
\qquad y COMP expr([ "+", "\\dfrac{" + expr([ "*", -A, "x" ]) + "}{" + B + "}", "\\dfrac{" + C + "}{" + B + "}" ])
\qquad y COMP \purple{PRETTY_SLOPE} \green{+ YINT}
The y-intercept is YINT
and the slope is decimalFraction( SLOPE, true, true )
.
Since the y-intercept is YINT
, the line must pass through the point (0, YINT)
.
The slope is decimalFraction( SLOPE, true, true )
. Remember that the slope tells you rise over run.
For every step we take we must move
abs( SLOPE_FRAC[0] )
positions
down (because it's negative)up.
For every step we take we must move
abs( SLOPE_FRAC[0] )
position
down (because it's negative)up.
For every step we take we must also move
SLOPE_FRAC[1]
positions
to the right.
For every step we take we must also move
SLOPE_FRAC[1]
position
to the right.
So the line must also pass through (SLOPE_FRAC[1], YINT + SLOPE_FRAC[0])
.
Since our inequality has a less-than or equal to sign, that means that any point below the line is a solution to the inequality, so the area below the line should be shaded.
Since our inequality has a greater-than or equal to sign, that means that any point above the line is a solution to the inequality, so the area above the line should be shaded.
Note that since the sign is less-than or equal to, any point on the line is also a solution, so the line should be solid. Note that since the sign is greater-than or equal to, any point on the line is also a solution, so the line should be solid.
Note that since the sign is less-than (and not equal to), any point on the line is not part of the solution, so the line should be dashed to indicate this. Note that since the sign is greater-than (and not equal to), any point on the line is not part of the solution, so the line should be dashed to indicate this.