How do you find the axis of symmetry, vertex and x intercepts for #y=1/2x^2-x-1#?

1 Answer
Dec 18, 2016

Axis of symmetry is #x=1#

#x_("intercepts")=0.73" and "-2.73# to 2 decimal places

#"Vertex "->(x,y)=(1,-3/2)#

Explanation:

#color(brown)("Note that the question does not state the method to be used.")#

#color(blue)("A quick trick to find the axis of symmetry.")#

#y=1/2(x^2-2x)-1#

Using the -2 from #-2x# we have:

#color(blue)(x_("vertex")=(-1/2)xx(-2) = +1) color(red)(larr"axis of symmetry " x=1)#
The above process is part of completing the square.

#color(blue)(y_("vertex") = 1/2(1)^2-(1)-1 = - 1 1/2->-3/2)#

#color(blue)(=>"Vertex "->(x,y)=(1,-3/2))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

To find the x-intercepts

Using the formula

#y=ax^2+bx+c" "=>" "x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=1/2; b=-1; c=-1#

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

#x=-1+-sqrt(3)#

#x=0.73205..." and "x=-2.73205...#

Tony B