what is the vertex?

Question

y=-2x²+8x-11

Answer 1

Hi Cassandra,

 

Well, there are two ways to go about this, and I don't know which one your book/teacher wants.  So, I'll give a few hints, and see if any of those help.

 

Method #1

 

Some textbooks give a direct formula for the vertex of a parabola, if you're given the equation in standard form.  That's what you have:  y = ax² + bx + c.  That's standard form.  The important numbers are the coefficients, so in your case, "a" is the number in front of the x² (including the negative sign), "b" is the number in front of the "x," and "c" is the constant at the end.  

 

The vertex of a parabola is usually given in terms of h (the x coordinate) and k (the y coordinate).  So, if you're looking for the point at (h, k), does your book give you a formula to figure out h & k based on the values of "a" "b" and "c"?  If so, plug 'em in, do the arithmetic, and poof, you have your vertex.

 

Method #2

 

Other textbooks don't give you a formula.  Instead, they expect you to take the standard form of an equation and convert it into vertex form:

 

4p(y - k) = (x - h)²

 

There are other ways of writing it, too, but the important parts are the (y - k) and the (x - h)².  The "h" and the "k" are the coordinates of the vertex...just like in Method #1.  The problem is that your parabola isn't written in that form!  :)  So, ya gotta convert it.

 

Without going too much further, the method is called "completing the square."  You have -2x² + 8x.  If you factor out the -2 in front of the x², you still have something with x squared, and then something times x.  The goal is to be able to write the equation in vertex form.  See the (x - h) thing up above?  It's squared.  If you remember factoring equations with an x squared in them, they sometimes end up being perfect squares...both factors are the same.  Like (x-2)(x-2).  That's (x-2)².  Fancy way of writing it, but they're the same thing. 

 

You need to find a number--a constant--that you can add to this situation so that, when you factor it, you get a perfect square, like (x - h)².  It's a quick little puzzle, and there's a shortcut to solve it.

 

And this is getting long.  :)  So, either you have a formula to figure out h and k from the original equation, or you have to complete the square, and then write the equation in vertex form...where you'll be able to see h and k pop out at you.

 

Does that help?  Let us know, and I'm sure either I or someone else can help take it the rest of the way...

 

-- Michael

Answer 2

what is the vertex?
y=-2x2+8x-11;I need to convert the equation to vertex form:

Y+11=-2(x^2 - 4x)

y+11-2( )= -2(x^2-4x) => y+11-2(4) = -2(x^2 -4x +4); Y+3=-2(x-2)^2

y= -2(x-2)^2 +(-3)

Then the vertex is located at h=2 and k=-3

(2,-3)

 

Answer 3

The answer is 4. thank you!