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

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