How do you find the vertex and the intercepts for #y = 1/4x^2 + 1/ 2x - 3/4#?

# y= 1/4 x^2+1/2 x -3/4 # . Comparing with standard equation

#y=ax^2+b+c # we get # a= 1/4 , b = 1/2 ,c = -3/4# We know

vertex # (x) = -b/(2a) = -1/2/(2*1/4) = -1#. Putting #x= -1#

in the equation # y= 1/4 x^2+1/2 x -3/4 #,

we get, vertex #(y) = 1/4 (-1)^2 +1/2(-1)-3/4 =1/4-1/2-3/4 #

#= -4/4= -1# .Vertex is at # (-1,-1) # . Putting # x=0#

in the equation we can get y intercept as #y= -3/4 or (0,-3/4)#,

putting # y=0# in the equation we can get x intercepts as

# 0= 1/4 x^2+1/2 x -3/4 or x^2+2x-3=0 #

or #(x+3)(x-1)=0 :. x = -3 or x = 1# , x intercepts are at

#(-3,0) and (1,0)#

graph{1/4x^2+1/2x-3/4 [-10, 10, -5, 5]}

Given: #y=1/4x^2+1/2x-3/4#

Compare to the form of #y=ax^2+bx+c#

#color(blue)("Determine the y intercept")#

#y_("intercept")=c=-3/4 ->(x,y)=(0,-3/4)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine the vertex")#

You have three methods:

1. Using the completed square form equation you can read it directly off but with a small amount of adjustment.
   Complete breakdown on this can be found on https://socratic.org/s/aHr8G5h4

2.factorising or use the formula to determine the #x# intercepts and the #x# value of the vertex will be half way between

1. You can do it this way:
   Note that this process is part of that for completing the square.

Write as:

#y=1/4(x^2color(red)(+2x))-3/4#

#x_("vertex") = (-1/2)xx(color(red)(+2)) = -1#

Buy substitution

#y_("vertex")=1/4(-1)^2+1/2(-1)-3/4 = -1#

Vertex#->(x,y)=(-1,-1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the "x_("interpts")#

Using other method
#y=ax^2+bx+c =0color(white)(..)# where #x=(-b+-sqrt(b^2-4ac))/(2a)#

#a=1/4; b=1/2; c= -3/4#

#x=(-1/2+-sqrt((1/2)^2-4(1/4)(-3/4)))/(2xx1/4)#

#x=-1+-2sqrt( 1/4+3/4)#

#x=-1+-2#

#x=-3 and x=1#