In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$y=x^2-2x-3$

The given information is:

$y=x^2-2x-3$

$$\begin{gathered} y=a{x}^2+bx+c \\ y=x^2-2x-3 \end{gathered}$$

The value of coefficient ais positive therefore we can say that the parabola opens up.

The axis of symmetry is the line $x=-\frac{b}{2a}$.

Substitute the values of a, and binto the equation.

Therefore, the axis of symmetry is $x=1$

The vertex is on the line of symmetry, so its x-coordinate will be $x=1$.

Now substitute the value of xinto the equation,

$$\begin{gathered} y={\left(1\right)}^2-2\left(1\right)-3 \\ =1-2-3 \\ =-4 \end{gathered}$$

Now substitute xequal to 0 in the equation to find the intercept y,

$$\begin{gathered} y=0^2-2\left(0\right)-3 \\ =0-0-3 \\ =-3 \end{gathered}$$

Therefore, the point $\left(0,-3\right)$ is the y-intercept.

We can see that the resultant point is left of the line of symmetry by 1 unit.

Therefore the point right to the line of symmetry by 1 unit is $\left(2,-3\right)$.

Now substitute yequal to 0 in the equation to find the intercept x,

$$\begin{gathered} x^2-2x-3=0 \\ \left(x-3\right)\left(x+1\right)=0 \\ x=-1,3 \end{gathered}$$

Therefore, the points of the x-intercept are $\left(-1,0\right)\mathrm{and}\left(3,0\right)$.