A anthonye Jul 2010 616 5 Apr 4, 2016 #1 Hi; I know the form y= mx + c, but is this the same y = m(x + constant) + c.

topsquark Forum Staff Jan 2006 11,578 3,454 Wellsville, NY Apr 4, 2016 #2 anthonye said: Hi; I know the form y= mx + c, but is this the same y = m(x + constant) + c. Click to expand... It's still a line, but with a different y intercept. \(\displaystyle y = m(x - x_0) + b = mx + ( -mx_0 + b)\) -Dan

anthonye said: Hi; I know the form y= mx + c, but is this the same y = m(x + constant) + c. Click to expand... It's still a line, but with a different y intercept. \(\displaystyle y = m(x - x_0) + b = mx + ( -mx_0 + b)\) -Dan

A anthonye Jul 2010 616 5 Apr 4, 2016 #3 Is it still called slope intercept form even with the brackets.

D Debsta MHF Helper Oct 2009 1,343 620 Brisbane Apr 4, 2016 #4 Slope-intercept form is y= mx + c because m is the slope and c is the y-intercept. y = m(x + constant) + c is not in slope-intercept form because although m is the slope, the y-intercept is not c.

Slope-intercept form is y= mx + c because m is the slope and c is the y-intercept. y = m(x + constant) + c is not in slope-intercept form because although m is the slope, the y-intercept is not c.