what is the vertex form of x^{2} + 5x + 6
i dont know how to do it
thank you
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).
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
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