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

Question

1596 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-21T22:51:49

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

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 $x = k = -2.5$

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

$h = 21$

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

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