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Calculus1 21 Online
OpenStudy (anonymous):

what would Y=-x^2+x+12 be in intercept form?

OpenStudy (amistre64):

intercept form ... thats a term you dont here often. I believe that form is similiar to teh structure of an ellipse of hyperbola

OpenStudy (anonymous):

great minds http://openstudy.com/study#/updates/51d43840e4b0befebdaf5489

OpenStudy (amistre64):

y=-x^2+x+12 x^2-x + y = 12 (x^2-x)/12 + y/12 = 1 and complete the square on the xs

OpenStudy (amistre64):

might be best to do a square completion before hand to adjust that 12 correctly

OpenStudy (anonymous):

oooh i see a quick google search tells me that "intercept form" is the same as "factored"

OpenStudy (amistre64):

the intercept form of a line :) ax + by = c ax/c + by/c = 1 x y ---- + ---- = 1 c/a c/b

OpenStudy (anonymous):

as in "display the \(x\) intercepts" i have never heard it called that before learn something new every day

OpenStudy (amistre64):

.... murmur

OpenStudy (anonymous):

lol

OpenStudy (austinl):

Not really on this topic, but I would like to at least get a color other than purple on here. Also, these guys are very helpful!

OpenStudy (amistre64):

lol, my sister says its brown :)

terenzreignz (terenzreignz):

amistre and satellite in one question... overkill :D

OpenStudy (austinl):

Exactly! So much intelligence in one small space.... universe... imploding... :D

OpenStudy (anonymous):

dsm, you can probably just ignore those first four posts, as I would imagine they're a bit confusing for you. The fourth post is spot on though. The "intercept form" is called the intercept form because, when written in that form, it is simple to look at the equation and see the x-intercepts. The form we'd like for this is factored. So, for this problem, we'd like to factor it. \(\Large -x^2 + x+12\) \(\Large = -(x^2 -x -12)\) \(\Large = -(x\pm\text{_})(x\pm\text{_})\)

OpenStudy (anonymous):

The first step is to get rid of the negative coefficient for \(x^2\) by factoring out a negative. The second step I've left for you to complete on your own. :)

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