Question

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

Answer 1

`vertex ➝ (1,-4)`

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

Explanation:

The vertex formula is:
`x_v=(-b)/(2a)`

so,
`x_v=(2)/(2*1)=1`

we substitute in the function equation,
`y_v=f(1)=1^2-2*1-3=-4`

so we get the point `(1,-4)`.
To know the x-axis interception points we equal the function to `0`.
`0=x^2-2x-3`

`x=(-b+-sqrt(b^2-4ac))/(2a)=(2+-sqrt((-2)^2-4*1*(-3)))/(2*1)`

`x_1=-1`
`x_2=3`

so we have the points `(-1,0)` and `(3,0)`.
If we substitute `x`by `0` in the function equation we get the interception point in the y-axis,
`f(0)=0^2-2*0-3=-3`

we get the point `(0,-3)`.