How do you find the vertex, axis of symmetry and intercepts of #f(x)=x^2+x+7#?

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Answer 1 · 2018-06-11T13:25:03

vertex #(-1/2,6 3/4)#

#x=-1/2# is the axis of symmetry

(0,7) is the intercept.

#(-b)/(2a)# gives the x coordinate for the vertex

#-1/2#

Put this into the equation

#y=1/4-1/2+7#

#y=6 3/4#

when x=0, y=7

when y=0, x=#[-1\pmsqrt(1-28)]/2#

#x=[-1\pmsqrt-27]/2# cannot have a negative square root so no solutions