randRangeNonZero( -3, 3 ) randRangeNonZero( -2, 2) randRange( 0, 4 ) (function() { var points = []; for ( var x = -2, i = 0; x <= 2; x++, i++ ) { var y = x * M + B, jitter = randRangeNonZero( -2, 2 ); if ( i === CORRECT ) { jitter = 0; } points.push( [ x, y + jitter ] ); } return points; })() function( arr ) { return "(" + arr.join( ", " ) + ")"; }

Which of the following ordered pairs represents a solution to the equation below?

$.map( POINTS, PAIR ).join( "\\qquad" )

y = expr(["+", ["*", M, "x"], B])

PAIR( POINTS[CORRECT] )

  • PAIR( point )

We can try plugging in the x-value of each ordered pair into the equation.

If we evaluate and get the y-value of the ordered pair, then that ordered pair is a solution!

Let's consider PAIR( point ).

If we plug in point[ 0 ] for x and evaluate, do we get point[ 1 ]?

y = (M)(point[ 0 ]) + B = M * point[ 0 ] + B = M * point[ 0 ] + B

Thus the only ordered pair that is a solution to the equation is PAIR( POINTS[ CORRECT ] ).

We come to the same answer by plotting the points and the equation.

graphInit({ range: 10, scale: 20, axisArrows: "<->", tickStep: 1, labelStep: 2 }); style({ stroke: BLUE, fill: BLUE }); plot(function( x ) { return x * M + B; }, [ -10, 10 ]); $.each( POINTS, function( i, point ) { if ( i === CORRECT ) { style({ stroke: ORANGE, fill: ORANGE }, function() { circle( point, 0.3 ); }); } else { circle( point, 0.15 ); } });

Which of the following ordered pairs represents a solution to the equation graphed below?

$.map( POINTS, PAIR ).join( "\\qquad" )

graphInit({ range: 10, scale: 20, axisArrows: "<->", tickStep: 1, labelStep: 2 }); style({ stroke: BLUE, fill: BLUE }); plot(function( x ) { return x * M + B; }, [ -10, 10 ]);

PAIR( POINTS[CORRECT] )

  • PAIR( point )

Let's try graphing each of the points.

circle( point, 0.15 );

The only point that falls on the line is PAIR( POINTS[ CORRECT ] ).

style({ stroke: ORANGE, fill: ORANGE }); circle( POINTS[ CORRECT ], 0.3);