Question

How do you find the vertex and the intercepts for `f(x)=x^2 - 12x -2`?

Answer 1

Vertex: `(6,-38)`

`x`-intercepts are:

`(6+sqrt38, 0)` and `(6-sqrt38, 0)`

`y`-intercept is:

`(0, -2)`

Explanation:

From this form of the parabola equation, i.e. `ax^2+bx+c`, the `x`-coordinate of the vertex is:

`x=(-b)/(2a)`

Here, `a=1` and `b=-12`, and `c` is `-2` therefore:

`x=(-(-12))/(2(1))=12/2=6`

Now we plug that into the equation to find `y`:

`y=6^2-12(6)-2=36-72-2=36-74=-38`

Vertex: `(6,-38)`

To find the `x`-intercepts, you have to set `y` (the equation) equal to `0` and solve for `x`

`x^2-12x-2=0`

We cannot factor this. Let's use the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-12)+-sqrt((-12)^2-4(1)(-2)))/(2(1)`

`x=(12+-sqrt(144+8))/2=(12+-sqrt152)/2=(12+-sqrt(2^2*38))/2=(12+-2sqrt38)/2=(2(6+-sqrt38))/2=(cancel2(6+-sqrt38))/cancel2`

`x=6+-sqrt38`

Therefore, the `x`-intercepts are:

`(6+sqrt38, 0)` and `(6-sqrt38, 0)`

To find the `y`-intercept, you have to plug in `0` for `x`:

`y=0^2-12(0)-2=-2`

Therefore, the `y`-intercept is:

`(0, -2)`