How do you factor #y=x^2 + 4x - 5 #?

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Answer 1 · 2016-03-18T19:07:50

If you mean to write it as a parabola in the form #y=a*(x-h)^2+k#

we have that

#y=(x^2+4x+4)-9=(x+2)^2-9#

If you mean to factor the LHS it is as follows

#y=x^2 + 4x - 5=>y=x^2+4x-4-1=> y=(x^2-1)+4(x-1)=> y=(x-1)(x+1)+4(x-1)=> y=(x-1)*(x+5)#

The graph of #y# is

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