# How do you find the vertex and the intercepts for y=x^2-2x-3?

Nov 23, 2017

vertex ➝ (1,-4)

x-ax is ➝ (-1,0) ∪ (3,0)
y-ax is ➝ (0,-3)

#### Explanation:

The vertex formula is:
${x}_{v} = \frac{- b}{2 a}$

so,
${x}_{v} = \frac{2}{2 \cdot 1} = 1$

we substitute in the function equation,
${y}_{v} = f \left(1\right) = {1}^{2} - 2 \cdot 1 - 3 = - 4$

so we get the point $\left(1 , - 4\right)$.
To know the x-axis interception points we equal the function to $0$.
$0 = {x}^{2} - 2 x - 3$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot 1 \cdot \left(- 3\right)}}{2 \cdot 1}$

${x}_{1} = - 1$
${x}_{2} = 3$

so we have the points $\left(- 1 , 0\right)$ and $\left(3 , 0\right)$.
If we substitute $x$by $0$ in the function equation we get the interception point in the y-axis,
$f \left(0\right) = {0}^{2} - 2 \cdot 0 - 3 = - 3$

we get the point $\left(0 , - 3\right)$.