What is the vertex form of #x = (2y +5 )^2 + 21#?

1 Answer
Aug 21, 2017

#x = 4(y - (-2.5))^2+ 21#

Explanation:

Given: #x = (2y +5 )^2 + 21#

Note: There is a quick way to do this but it is easy to confuse yourself so I will do it the following way.

Expand the square:

#x = 4y^2+20y+25+21#

#x = 4y^2+20y+46" [1]"#

This is the standard form

#x = ay^2 +by + c#

where #a = 4, b = 20 and c = 46#

The general vertex form is:

#x = a(y - k)^2+ h" [2]"#

We know that #a# in the vertex form is the same as #a# in the standard form:

#x = 4(y - k)^2+ h" [2.1]"#

To find the value of k, use the formula:

#k = -b/(2a)#

#k = -20/(2(4)) = -2.5#

#x = 4(y - (-2.5))^2+ h" [2.2]"#

To find h, evaluate equation [1] at #x = k = -2.5#

#h = 4(-2.5)^2+20(-2.5)+46#

#h = 21#

#x = 4(y - (-2.5))^2+ 21" [2.3]"#