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

Apr 10, 2016

Axis of Symmetry at: $x = 2$
Vertex at: $\left(2 , - 7\right)$

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:
$y = a {x}^{2} + b x + c$

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

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:
$T {P}_{x} = - \frac{b}{2 a}$
to determine the $x$ value of the vertex.
Substituting the appropriate values into the formula we get:

$T {P}_{x} = - \left(- \frac{4}{2 \cdot 1}\right)$

$= \frac{4}{2}$

$= 2$

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.

$y = {x}^{2} - 4 x - 3$
$y = {2}^{2} - 4 \cdot 2 - 3$
$y = - 7$

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 $\left(2 , - 7\right)$.

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 $x = 2$.