How do you find the vertex and the intercepts for #Y=x^2+5x-6#?

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Answer 1 · 2017-03-09T12:30:00

intercepts: # x = -6, 1#

vertex: #(- 5/2, - 49/4)#.

#y=x^2+5x-6#

For the x-axis intercepts, factor and set to zero:

#y =(x+6)(x-1) = 0#

#implies x = -6, 1#

For the vertex, complete the square:

#y=(x+5/2)^2 - 25/4 -6#

#=(x+5/2)^2 - 49/4 #

So this quadratic has its minimum value at #(- 5/2, - 49/4)#.