What is the vertex form of #7y = 19x^2+18x+42#?

1 Answer
May 25, 2017

#y=19/7(x+9/19)^2+717/133#

Explanation:

Strategy: Use the technique of completing the square to put this equation in vertex form:

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

The vertex can be pulled from this form as #(h,k)#.

Step 1. Divide both sides of the equation by 7, to get #y# alone.

#y=19/7 x^2+18/7 x+ 6#

Step 2. Factor out #19/7# to get #x^2# alone.

#y=19/7(x^2+7/19xx18/7+7/19xx6)#

Notice we just multiply each term by the reciprocal to factor it out.

Step 3. Simplify your terms

#y=19/7(x^2+18/19x+42/19)#

Step 4. For the term in front of #x#, you must do three things. Cut it in half. Square the result. Add and subtract it at the same time.
Term next to #x#: #18/19#
Cut it in half: #1/2xx18/19=9/19#
Square the result: #(9/19)^2=81/361#
Finally, add and subtract that term inside the parenthesis:

#y=19/7(x^2+18/19x+color(red)(81/361)-color(red)(81/361)+42/19)#

The part that can now be expressed as a perfect square is in blue.

#y=19/7(color(blue)(x^2+18/19x+81/361)-81/361+42/19)#

This gives you the perfect square using the number you got when you cut it in half (i.e., #9//19#)

#y=19/7(color(blue)((x+9/19)^2)-81/361+42/19)#

Combine the remaining two fractions inside the parenthesis.

#y=19/7((x+9/19)^2+717/361)#

Step 5. Multiply the #19/7# back through to each term.

ANSWER: #y=19/7(x+9/19)^2+717/133#

So the vertex is at #h=-9/19# and #k=717/133# which can be expressed as

#(-9/19, 717/133)~~(0.4737,5.3910)#