How do you write the equation in vertex form #x=3y^2+5y-9#?

1 Answer
Jan 5, 2017

Answer:

Please see the explanation.

Explanation:

The vertex form of a parabola of this type is:

#x = a(y - k)^2 + h" [1]"#

where, x and y correspond to any point, #(x, y)#, on parabola, h and k correspond to the point that is the vertex, #(h, k)#, and a is the leading coefficient of the #y^2# term in standard form.

Given: #x = 3y^2 + 5y + 9" [2]"#

Let's expand the square in equation [1], multiply the terms by a, and write it directly above equation [2]:

#x = ay^2 - 2aky + ak^2 + h" [1a]"#
#x = 3y^2 + 5y + 9" [2]"#

Please observe that #a = 3# so lets add zero to equation [2] in the form #3k^2 - 3k^2#:

#x = ay^2 - 2aky + ak^2 + h" [1a]"#
#x = 3y^2 + 5y + 3k^2 - 3k^2 + 9" [2a]"#

If we set second term in equation [1a] equal to the second term in equation [2a] and we substitute 3 for a, then we can solve for the value of k:

#-2aky = 5y#

Substitute 3 for a:

#-2(3)ky = 5y#

#k = -5/6#

This tells us that the first three terms in equation [2a] can be replaced by #3(y - -5/6)^2#:

#x = 3(y - -5/6)^2 - 3k^2 + 9" [2b]"#

Substitute #(-5/6)# for k:

#x = 3(y - -5/6)^2 - 3(-5/6)^2 + 9" [2c]"#

Simplify:

#x = 3(y - -5/6)^2 + 83/12" [2d]"#

Equation [2d] is the vertex form for a parabola that opens to the right or left.

You can read that the vertex is at the point #(83/12, -5/6)# from equation [2d].