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

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