What are the x-intercept(s) of the graph of #y + 12 = x^2 + x#?

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Answer 1 · 2017-06-11T21:41:57

See a solution process below:

To find the #x#-intercepts we need to set #y# to #0# and solve for #x#:

#y + 12 = x^2 + x# becomes:

#0 + 12 = x^2 + x#

#12 - color(red)(12) = x^2 + x - color(red)(12)#

#0 = x^2 + x - 12#

#0 = (x + 4)(x - 3)#

Solution 1)

#x + 4 = 0#

#x + 4 - color(red)(4) = 0 - color(red)(4)#

#x + 0 = -4#

#x = -4#

Solution 2)

#x - 3 = 0#

#x - 3 + color(red)(3) = 0 + color(red)(3)#

#x - 0 = 3#

#x = 3#

The #x#-intercepts are: #-4# and #3#

Or

#(-4, 0)# and #(3, 0)#