# For the function y = x^2 - 2x - 5, determine: a) Its vertex b) The x-intercepts in simplified form.

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Student Comments

aruv | Student

`y=x^2-2x-5`

`y=x^2-2x+1-1-5`

`y=(x-1)^2-6`

`y+6=(x-1)^2`

So the vertex of the above parabola (function) is

y+6=0

y=-6

and

x-1=0

x=1

Thus coordinate of the vertex is (1,-6)

The x-intercepts , we can calculate by substituting y=0 in the given function

0+6=(x-1)^2

`(x-1)^2=6`

`(x-1)=+-sqrt(6)`

`x=1+-sqrt(6)`

Thus x intercepts are `1+sqrt(6) and 1-sqrt(6).`

Now we can see above results graphically as

Where graph are intersecting x-axis. These are between 3 & 4 , and

-2 and -1.