Find the vertex, axis of symmetry, and graph the parabola? x = 5y^2 -20y +23

1 Answer
Oct 20, 2017

Please see below.

Explanation:

This function is the quadratic function of the type #xf(y)=ay^2+by+c#. This when written in the form #x=a(y-k)^2+h# has #(h,k)# as vertex and #y-k=0# i.e. #y=k# as axis of symmetry.

To draw the graph, we select a few values of #y# around #k# i.e. both sides of #k# - less than this as well as greater than this, to find corresponding values of #x# and then draw graph.

Here we have #x=5y^2-20y+23#

= #5(y^2-4y)+23# and completing square

= #5(y^2-4y+4)-5xx4+23#

= #5(y-2)^2+3#

Hence vertex is #(3,2)# and axis of symmetry is #y-2=0# or #y=2#.

Selecting values #y=-6,-4,-2,0,2,4,6,8,10# and putting them in #x=5(y-2)^2+3#, we get #x=323,183,83,23,3,23,83,183,323#and graph drawn to scale appears as

graph{5(y-2)^2+3-x=0 [-3.26, 76.74, -16.84, 23.16]}

However it will look better when #x#-axis is compressed (i.e.not drawn to scale) as follows

graph{5(y-2)^2+3-x=0 [-10, 350, -10, 10]}