How do you find the vertex and the intercepts for #y = x^2 – 4x + 9#?

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Answer 1 · 2018-05-08T07:10:47

vertex is (2,5)
(0,9) is the y intercept
no x intercepts

Complete the square

#y=(x-2)^2-4+9#

#y=(x-2)^2+5#

so the vertex is (2,5)

If #y=x^2-4x+9# when #x=0, y=9#

If we take the equation after we have completed the square

#y=(x-2)^2+5# and put #y=0#

#0=(x-2)^2+5#

#(x-2)^2=-5#

#x-2=sqrt(-5)#

As we cannot take the square root of a negative number there are no roots so it does not cross the #y# axis