# How do you find the important parts of the equation to graph the function  f(x) = x² - 2x -3?

Apr 4, 2018

The coordinates of the vertex are at $\left(1 , - 4\right)$, the x-intercepts are at $x = - 1$, and $x = 3$, the y-intercept is at $y = - 3$.

#### Explanation:

This problem nicely demonstrates the advantages of being able to view a quadratic function in STANDARD, VERTEX, and FACTORED form.

The function as given is in STANDARD form.

$f \left(x\right) = {x}^{2} - 2 x - 3$

The standard form explicitly shows the y-intercept for the function.

If $f \left(x\right) = a {x}^{2} + b x + c$, then the y-intercept is at $\left(0 , c\right)$. In this case, $c = - 3$, so the y-intercept is at $\left(0 , - 3\right)$.

The standard form also shows whether the parabola is downward-facing or upward-facing. If $a$ is positive, then the parabola is upward-facing. Here $a = 1$ is positive, so the parabola is upward facing.

Next, we can convert this function to its vertex form by completing the square. The advantage of the vertex form is that it explicitly shows you the coordinates of the vertex. If we have a parabola in vertex form

$f \left(x\right) = a {\left(x - k\right)}^{2} + h$

then the coordinates of the vertex will be at $\left(k , h\right)$.

For our function,

$f \left(x\right) = {x}^{2} - 2 x - 3 = {x}^{2} - 2 x + 1 - 4 = {\left(x - 1\right)}^{2} - 4$, so the coordinates of the vertex are at $\left(1 , - 4\right)$.

Finally we want to look at the function in factored form so that we can locate its x-intercepts if any exist. If a function is in factored form

$f \left(x\right) = a \left(x - {r}_{1}\right) \left(x - {r}_{2}\right)$

and the x-intercepts will be at $x = {r}_{1}$ and $x = {r}_{2}$.

The factored form for this function is

$f \left(x\right) = \left(x + 1\right) \left(x - 3\right)$

so the x-intercepts are at $x = - 1$, and $x = 3$.

We can put all of this information together to sketch the graph of our function.

graph{x^2-2x-3 [-5, 5, -5, 5]}