y=-(x+3) squared
what is the vertex?
2006-11-13 08:01:39 · 4 answers · asked by HOME SLICE! 2 in Science & Mathematics ➔ Mathematics
how do you get the vertex?
2006-11-13 08:02:28 · update #1
there is a vertex, i know its a parabala(im not sure thats how you spell it) i got a vertex for every other question on my math homework except that one
2006-11-13 08:10:12 · update #2
y=(x+3)^2=x^2+6x+9
y'=2x+6=0
2x=-6
x=-3
y(-3+3)^2=0
so the vertex is (-3,0)
2006-11-13 08:12:37 · answer #1 · answered by yupchagee 7 · 16⤊ 0⤋
I will assume that you dont know calculus. We need to find the maximum or minimum for the function. Fortunately, this is pretty easy. The minimum value for a square is zero. This occurs at (x+3) = 0 or x = -3. The value of the "equation" at x = -3 is 0. So the vertex occurs at (x,y) = (-3,0)
Since all the other values of this are negative, this is a maximum.
2006-11-13 16:21:53 · answer #2 · answered by heartsensei 4 · 0⤊ 0⤋
The smallest that (x+3)^2 can be is zero at x = -3, so that is the vertex.
There's no need to factor or anything. Note that the parabola is symmetric about x = -3, for example it has the same value at x= -3-1 as it does at x= -3+1
2006-11-13 16:23:57 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
There is none. This equation is for a straight line. Vertex is only for quadratic equations.
2006-11-13 16:06:28 · answer #4 · answered by Doug k 3 · 0⤊ 1⤋