Question

What is the axis of symmetry and vertex for the graph `y=x^2-4x-3`?

Answer 1

Axis of Symmetry at: `x=2`
Vertex at: `(2,-7)`

Explanation:

Note: I'll use the terms Turning Point and Vertex interchangeably as they are the same thing.

Let's first have a look at the vertex of the function

Consider the general form of a parabolic function:
`y=ax^2+bx+c`

If we compare the equation that you have presented:
`y=x^2-4x-3`

We can see that:
The `x^2` coefficient is 1; this implies that `a` = 1
The `x` coefficient is -4; this implies that
`b` = -4
The constant term is -3; this implies that `c` = 3
Therefore, we can use the formula:
`TP_x=-b/(2a)`
to determine the `x` value of the vertex.
Substituting the appropriate values into the formula we get:

`TP_x=-(-4/(2*1))`

`=4/2`

`=2`

Therefore, the `x` value of the vertex is present at `x=2`.

Substitute `x=2` into the given equation to determine the `y` value of the vertex.

`y=x^2-4x-3`
`y=2^2-4*2-3`
`y=-7`

Therefore, the `y` value of the vertex is present at `y=-7`.

From both the `x` and `y` values of the we can determine that the vertex is present at the point `(2,-7)`.

Now let's have a look at the function's Axis of Symmetry:

The axis of symmetry is essentially the `x` value of the turning point (the vertex) of a parabola.

If we have determined the `x` value of the turning point as `x=2`, we can then say that the axis of symmetry of the function is present at `x=2`.