How do you find the x and y intercepts for $x^2-6x+1$ ?

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Answer 1 · 2016-05-17T13:12:14

Intercepts on $x$ axis are $(3+2sqrt2,0)$ and $(3-2sqrt2,0)$

and intercept on $y$ axis is $(0,1)$ .

For finding $x$ and $y$ intercepts of $y=x^2-6x+1$

Let us first put $y=0$ (to find intercepts on $x$ axis)

or $x^2-6x+1=0$ .

Using quadratic formula $(-b+-sqrt(b^2-4ac))/(2a)$ , we get

$x=(-(-6)+-sqrt((-6)^2-4*1*1))/2=(6+-sqrt32)/2=3+-2sqrt2$

Hence, intercepts on $x$ axis are $(3+2sqrt2,0)$ and $(3-2sqrt2,0)$

For intercepts on $y$ axis, put $x=0$ i.e.

$y=0^2-6*0+1=1$

Hence, intercept on $y$ axis is $(0,1)$ .

graph{x^2-6x+1 [-8.5, 11.5, -5.64, 4.36]}

How do you find the x and y intercepts for x^2-6x+1x^2-6x+1?