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=323,183,83,23,3,23,83,183,323and 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]}