Question

How do you find the vertex and intercepts for `y=x^2+2x-1`?

Answer 1

Vertex`->(x,y)-> (-1,-2)`
x-axis intercepts `->x~~0.414" or "x~~-2.414` to 3 decimal places
y-axis intercept `=-1`

Explanation:

Given: `y=x^2+2x-1` ....................................(1)

`color(blue)("To determine the vertex.")`

This is already in the form of `y=a(x^2+b/a x-1)` because a=1

`x_("vertex")=(-1/2)xx b/a`
This is a variation on 'Vertex Form Equation'
'::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

In your case:
`color(blue)(x_("vertex")=(-1/2)xx(+2)=-1)`......................(2)

Substitute (2) into equation (1)

`y_("vertex")=(-1)^2+2(-1)-1`

`color(blue)(y_("vertex")=+1-2-1=-2)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To determine "y_("intercept")`

The graph crosses the y-axis when `x=0`

So `y=x^2+2x-1 -> y_("intercept")=(0)^2+2(0)-1`

`color(blue)(y_("intercept")=-1`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To determine "x_("intercepts")`

The graph crosses the x-axis when `y=0`

So `y=0=x^2+2x-1`

We need the form `y=0=(x+?)(x+?)` to find any integer factors

The only factor of 1 is 1 and to get -1 we need `(-1)xx(+1)`
which is fine until you try and get `+2x`. It fails then! So it means that the factors are not integer values. In which case we need to use the formula.
Using: `color(white)(......)y=ax^2+bx+c`

Where `color(white)(...)x=(-b+-sqrt(b^2-4ac))/(2a)`

`a=1 ; b=+2 ; c=-1`

`x=(-(+2)+-sqrt((+2)^2-4(+1)(-1)))/(2(1))`

`x=(-2+-sqrt(4+4))/2`

but `sqrt(8)=sqrt(2xx2^2)=2sqrt(2)`

`x=(-2+-2sqrt(2))/2`

`x~~0.414" or "x~~-2.414` to 3 decimal places