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

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Answer 1 · 2015-11-20T18:55:29

The x-coordinate of the vertex #=-b/(2a)#

Remember, the general form of a quadratic is:

#ax^2+bx+c=0#

x-coordinate of vertex #=(-b)/(2a)=(-(-1))/(2xx1)=1/2#

So, when #x=1/2#, then #y=(1/2)^2-(1/2)+2=7/4#

VERTEX #= (1/2, 7/4)#

Solve for #y=0# to find the y-intercepts :

#(x-2)(x+1)=0#

So, #x=2# and #x=-1#

The x-intercept is when #x=0#

#y=(0)^2-(0)+2=2#