Question

How do you find the vertex and the intercepts for `f(x)=x^2-6x-6`?

Answer 1

vertex: `(3,-15)`
y-intercept: `(-6)`
x-intercepts: `(3+sqrt(15))` and `(3-sqrt(15))`

Explanation:

Given
`color(white)("XXX")f(x)=x^2-6x-6`
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We can convert this into the general vertex form `(x-color(red)a)^2+color(blue)b` with vertex at `(color(red)a,color(blue)b)`
by "completing the square" methodology:

`x^2-6x-6`
`color(white)("XXX")=x^2-6xcolor(magenta)(+3^2)-6color(magenta)(-3^2)`
`color(white)("XXX")=(x-color(red)3)^2+color(blue)((-15))`
which is the vertex form with vertex at `(color(red)3,color(blue)(-15))`

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The y-intercept (or equivalently the `f(x)`-intercept)
is the value of `y` (or equivalently of `f(x)`) when `x=0`
`color(white)("XXX")f(0)=0^2-6 * 0 -6= -6`
So the y (or `f(x)`) intercept is at `y(=f(x))=-6`

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The x-intercepts are the values of `x` fro which `f(x)` (or equivalently, `y`) are equal to 0.
So we want the values of `x` for which
`color(white)("XXX")x^2-6x-6=0`
Since there is not obvious factoring of the left side,
we will use the quadratic formula to find the "roots" of this expression.
`color(white)("XXX")`For the general form `ax^2+bx+c=0`
`color(white)("XXX")`the quadratic formula tells us that the roots are
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`
In this case `a=1, b= -6, and c=-6`
so we have
`color(white)("XXX")x=(6+-sqrt((-6)^2-4 * 1 * (-6)))/(2 * 1)`

`color(white)("XXXX")=(6+-sqrt(36+24))/2`

`color(white)("XXXX")=(6+-sqrt(60))/2`

`color(white)("XXXX")=(6+-2sqrt(15))/2`

`color(white)("XXXX")=3+-sqrt(15)`

So the `x`-intercepts are at `3+sqrt(15)` and `3-sqrt(15)`

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