A Parabola with Vertex (0,4) and zeros -6 and 6 in factored form would be y=1/8(x-6)(x+6) and would open up right?...
So..what would the vertex form of this equation be..and what would i need to obtain it?
2007-04-12 12:17:42 · 2 answers · asked by J-Rome 2 in Science & Mathematics ➔ Mathematics
k yeh and that opens down which it should..so what did i do wrong in my factored equation to get a parabola that opened up?
2007-04-12 12:28:29 · update #1
what would be the procedure for factored form :(
2007-04-12 12:32:00 · update #2
y = (-1/9)x^2 + 4
In general,
y = c(x - a)^2 + b is a parabola with vertex at (a, b) . This is obtained by completing the square.
Note: Your answer y=1/8(x-6)(x+6) is not correct. The intercepts are 6 and -6, but when x = 0 you get -36/8 which is not equal to 4. The number in front should be -1/9 and not 1/8.
2007-04-12 12:23:13 · answer #1 · answered by snpr1995 3 · 0⤊ 0⤋
If the vertex is (p,q) the vertex equation would be
y=a(x-p)^2+q
You already have p and q
y=ax^2+4 As (6,0) belongs to the parabola
0=36a+4 so a= -1/9
y=-1/9x^2+4
2007-04-12 19:26:00 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋