# vertex form

• Jul 25th 2013, 07:45 PM
Mathnood768
vertex form
what is the vertex form of x2 + 5x + 6

i dont know how to do it

thank you
• Jul 25th 2013, 08:02 PM
MarkFL
Re: vertex form
You need to complete the square...are you familiar with this method?
• Jul 25th 2013, 09:10 PM
Mathnood768
Re: vertex form
is it...

(x^2 + 5x + 6.25)+6 - 6.25

i do not know how to make (x^2 + 5x + 6.25) factored
• Jul 26th 2013, 07:18 AM
HallsofIvy
Re: vertex form
Then where did you get that formula? In order to be able to do this problem, you have to know how to complete the square- you got 6.25, I presume, as $\displaystyle (5/2)^2$. Do you not know why you did that or what "completing the square" means?

The whole point is that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$. Comparing $\displaystyle x^2+ 2ax$ with $\displaystyle x^2+ 5x$, we must have $\displaystyle 2a= 5$ or $\displaystyle a= \frac{5}{2}= 2.5$. Then $\displaystyle a^2= (2.5)^2= 6.25$. That is, $\displaystyle (x+ 2.5)^2= x^2+ 5x+ 6.25$. So adding and subtracting 6.25 from both sides, $\displaystyle x^2+ 5x+ 6.25)+ 6- 6.25= (x+ 2.5)^2- .25$ which I would have preferred to write as $\displaystyle (x+ \frac{5}{2})^2- \frac{1}{4}$.

Now, you know (I hope!) that a square is never negative. When x= -5/2, x+ 5/2= 0 so $\displaystyle y= (x+ 5/2)^2- 1/4$ has value -1/4 while if x is any other number, $\displaystyle (x+ 5/2)^2- 1/4$ is larger than -1/4. The lowest point on the graph is (-5/2, -1/4).
• Jul 26th 2013, 03:27 PM
Mathnood768
Re: vertex form
i was actually asked to find the x and y intercepts and graph y = |x2 + 5x + 6|

i was trying to get it into vertex form so i could get the vertex point so i know how high the curve would be i still dont know, i think x = -5/2 is right from looking at the answers but the y intercept is 6, not -1./4

all i know is the y intercept is 6
and x intercepts are (-3, 0) and (-2, 0)
the vertex point is
• Jul 26th 2013, 11:50 PM
muddywaters
Re: vertex form
Quote:

Originally Posted by Mathnood768
i think x = -5/2 is right from looking at the answers but the y intercept is 6, not -1./4

the mistake is that the y-intercept is not at x=-5/2, y-intercept is always at x=0
x=-5/2 is where the vertex is, the vertex is what u get from completing the square. then the vertex is (-5/2, -1/4)

Edit:
but since it's a graph of y=|x^2+5x+6|, taking the absolute value, the "vertex" is (-5/2, 1/4). i think you should try to draw the graph, and you'll understand it better