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Mathematics 14 Online
OpenStudy (anonymous):

What are the x-intercept(s) of the graph of y + 12 = x2 + x?

OpenStudy (anonymous):

in order to find x-intercept(s) simply put y=0 and then solve for x in the resultant equation

OpenStudy (anonymous):

0+12=x^2+x ?

OpenStudy (anonymous):

y + 12 = x2 + x putting y=0 we have 12 = x2 + x x^2 +x-12=0 x^2+4x -3x -12 =0 x(x+4)-3(x+4)=0 (x+4)(x-3)=0 hence x=3,-4 and the corresponding point is (3,0) and (-4,0)

OpenStudy (anonymous):

ok so i insert 0 instead of y, is there a formula that I have to use or something?

OpenStudy (precal):

After you sub y=0, then you have to look at the quadratic and use an appropriate method to solve it. You can use factoring, factoring by grouping is what matricked used, or the quadratic formula. X-intercepts are the location where the function crosses the x-axis.

OpenStudy (anonymous):

thank you!

OpenStudy (precal):

yw

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