How do you find the vertex and the intercepts for #y = x^2 - 6x + 11#?

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Answer 1 · 2018-05-16T07:22:05

y intercept (0,11)
vertex (3,2)
does not cross the x axis

The constant is 11 so the y intercept is (0,11)

complete the square

let #x^2-6x+11=0#

#(x-3)^2-9+11=0#

#(x-3)^2+2=0#

so the x coordinate of the vertex is 3.
Put x=3 into the equation

#y=3^2-6xx3+11#

#y=9-18+11#

#y=2#

So (3,2) is the vertex.

As it is a positive quadratic it is #uu# shaped so the vertex is a minimum. As the minimum is above the x axis it does not cross the x axis