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

1 Answer
Jan 19, 2016

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

Tony B