Expert: Paul Klarreich - 2/25/2008

Question
"I understand what a vertex is and how to find it on a graph or a graphing calculator but, do not know how to do it using an equation. The question is find the vertex of y=2x^2 +11x-6 I already know what the vertex is for the equation but, do not know how to show work for it without saying I used the calculator. I also know how to use the equation b/(-2a) but, when using it for the equation the x I get holds to be correct but, when I plug it into the equation to get the y the y turns out to become a number in the 50's instead of -21 as it should be. I am not sure if I am plugging it in wrong or if the equation just does not hold true for all problems. More info.: I used the equation b/(-2a) to find the x coordinate and got -2.75 which checks with the graph. So theoretically if I have the x coordinate I should just be able to plug in the x into the equation y=2x^2 +11x-6 which would be y=2 * -2.75^2 + 11 * -2.75 - 6 then I found y to be -51.375 but, I know that it is incorrect because y should be -21. I also can not use the completing the square method so am trying to find an alternative route ergo b/(-2a). Sarah"

We are doing graphing problems as a review to make sure all students in the class are on the same page. I currently take pre-calculus and in class we are learning to add, subtract and multiply matrices and find the determinants of matrices.

Answer
Questioner:   Sarah
Category:  Calculus
Private:  No
 
Subject:  How to find the vertex of a parabola?
Question:  "I understand what a vertex is and how to find it on a graph or a graphing calculator but, do not know how to do it using an equation. The question is find the vertex of y=2x^2 +11x-6 I already know what the vertex is for the equation but, do not know how to show work for it without saying I used the calculator. I also know how to use the equation b/(-2a)
>> Better is to write it as  -b/2a.

but, when using it for the equation the x I get holds to be correct

>> Was it  x = -11/4 ?

but, when I plug it into the equation to get the y the y turns out to become a number in the 50's instead of -21 as it should be.

>> Was it:  

y = 2x^2 + 11x - 6

y = 2(-11/4)^2 + 11(-11/4) - 6

y = 2(121/16) - 121/4 - 6

y = 121/8 - 121/4 - 6

y = - 121/8  - 48/8

y = - 169/8

y = -21 and 1/8

Does that answer your question?




I am not sure if I am plugging it in wrong or if the equation just does not hold true for all problems.

>> Please!  

More info.: I used the equation b/(-2a) to find the x coordinate and got -2.75 which checks with the graph. So theoretically if I have the x coordinate I should just be able to plug in the x into the equation y=2x^2 +11x-6 which would be y=2 * -2.75^2 + 11 * -2.75 - 6 then I found y to be -51.375 but, I know that it is incorrect because y should be -21.

>> No, it's incorrect because you are not substituting correctly.  Go home and write, using PEN AND INK:

I WILL ALWAYS USE PARENTHESES WHEN SUBSTITUTING.

100 times.

And don't use decimal forms for this.  Fractions work much better, and you will do much better work when you become good at using them.

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I also can not use the completing the square method

>> Why not?  That should be your favorite.

so am trying to find an alternative route ergo b/(-2a). Sarah"

We are doing graphing problems as a review to make sure all students in the class are on the same page. I currently take pre-calculus and in class we are learning to add, subtract and multiply matrices and find the determinants of matrices.

Try these links to previous questions:

http://en.allexperts.com/q/Advanced-Math-1363/2008/2/Parabola-2.htm

http://en.allexperts.com/q/Advanced-Math-1363/2008/2/Parabola-3.htm