what is the vertex form of x^{2}+ 5x + 6

i dont know how to do it

thank you

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- Jul 25th 2013, 08:45 PMMathnood768vertex form
what is the vertex form of x

^{2}+ 5x + 6

i dont know how to do it

thank you - Jul 25th 2013, 09:02 PMMarkFLRe: vertex form
You need to complete the square...are you familiar with this method?

- Jul 25th 2013, 10:10 PMMathnood768Re: 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, 08:18 AMHallsofIvyRe: 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 . Do you not know

**why**you did that or what "completing the square"**means**?

The whole point is that . Comparing with , we must have or . Then . That is, . So adding and subtracting 6.25 from both sides, which I would have preferred to write as .

Now, you know (I hope!) that a square is never negative. When x= -5/2, x+ 5/2= 0 so has value -1/4 while if x is any other number, is**larger**than -1/4. The lowest point on the graph is (-5/2, -1/4). - Jul 26th 2013, 04:27 PMMathnood768Re: 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 27th 2013, 12:50 AMmuddywatersRe: vertex form
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