# How do you find the important points to graph y=x^2-2x-3?

The important points are x-intercepts $\left(- 1 , 0\right)$ and $\left(3 , 0\right)$, y-intercept $\left(0 , - 3\right)$, and the Vertex $\left(1 , - 4\right)$

#### Explanation:

From the given equation $y = {x}^{2} - 2 x - 3$ , set $x = 0$ then solve for the y-intercept

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

$y = {0}^{2} - 2 \cdot 0 - 3$

$y = - 3$

From the given equation $y = {x}^{2} - 2 x - 3$ , set $y = 0$ then solve for the x-intercept

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

${x}^{2} - 2 x - 3 = 0$

by factoring method

$\left(x - 3\right) \left(x + 1\right) = 0$

$x = 3$ and $x = - 1$ when $y = 0$

so $\left(3 , 0\right)$ and $\left(- 1 , 0\right)$ are x-intercepts

From the given equation $y = {x}^{2} - 2 x - 3$ ,by completing the square, find the vertex

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

$y = {x}^{2} - 2 x + 1 - 1 - 3$

$y = {\left(x - 1\right)}^{2} - 4$

$y - - 4 = {\left(x - 1\right)}^{2}$

the vertex (h, k)=$\left(1 , - 4\right)$

graph{y=x^2-2x-3[-20,20, -10, 10]}

have a nice day from the Philippines

Feb 8, 2016

Axis of symmetry: $x = 1$
Vertex: $\left(1 , - 4\right)$
X-intercepts:$\left(- 1 , 0\right)$ and $\left(3 , 0\right)$

#### Explanation:

$y = {x}^{2} - 2 x - 3$ is a quadratic equation in standard form, $a x + b x + c$, where $a = 1 , b = - 2 , c = - 3$. The graph of a quadratic equation is a parabola.

You need the axis of symmetry, the vertex, and the x-intercepts.

Axis of Symmetry
The axis of symmetry is an imaginary line dividing the parabola into two equal halves. The formula for the axis of symmetry is $x = \frac{- b}{2 a}$.

Substitute the given values for $a$ and $b$ into the formula for the axis of symmetry.

$x = \frac{- b}{2 a}$

$x = \left(- \frac{- 2}{2 \cdot 1}\right) =$

$x = \frac{2}{2} =$

$\textcolor{g r e e n}{x = 1}$

This is the axis of symmetry, and it is also the $x$ value for the vertex.

Vertex
The vertex is the maximum or minimum point of a parabola. Since $a > 0$, the vertex is the minimum and the parabola opens upward.

Substitute $1$ for $x$ into the quadratic equation and solve for $y$.

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

$y = {1}^{2} - \left(2 \cdot 1\right) - 3 =$

$y = 1 - 2 - 3$

$\textcolor{p u r p \le}{y = - 4}$

The vertex is color(green)((1,color(purple)(-4)).

X-Intercepts
The x-intercepts are the values of $x$ that intersect the y-axis. A parabola has two x-intercepts.

Substitute $0$ for $y$ in the quadratic equation.

$0 = {x}^{2} - 2 x - 3$

Factor ${x}^{2} - 2 x - 3$

Find two numbers that when added equal $- 2$, and when multiplied equal $- 3$. The numbers $1$ and $- 3$ fit the pattern. Rewrite the equation as its factors.

$\textcolor{red}{\left(x + 1\right)} \textcolor{b l u e}{\left(x - 3\right)} = 0$

First solve $\textcolor{red}{\left(x + 1\right)} = 0$

Subtract $\textcolor{red}{1}$ from both sides.

$\textcolor{red}{x = - 1}$

Next solve $\textcolor{b l u e}{\left(x - 3\right)} = 0$

Add $\textcolor{b l u e}{3}$ to both sides.

$\textcolor{b l u e}{x = 3}$

The x-intercepts are $\left(- 1 , 0\right)$ and $\left(3 , 0\right)$.

graph{y=x^2-2x-3 [-10, 10, -5, 5]}