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]}