i am looking at a graph which is almost like -(x-1)^2+1 except that the zeros are at -1 and at 2 instead of at 0 and 2. how could i change the equation so that this touches the axis at -1 and 2?
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so assuming that the vertex is the exact same, we have (1, 1) as our vertex.
BUT use factored form
we know that
x = -1 and x = 2
x+1 = 0 and x - 2 = 0
SO our equation in factored form is
y = a(x+1)(x-2)
and sub vertex in for x and y to solve for a
1 = a(1+1)(1-2)
1 = a(2)(-1)
-2a = 1
a = -1/2
THEREFORE our equation is y = -1/2(x+1)(x-2)
y = -1/2(x^2 +x -2x -2)
y = -1/2(x^2 -x - 2)
y = -1/2x^2 + 1/2x + 1
which if you want to get it into vertex form
y = -1/2(x^2 - x) + 1
y = -1/2(x^2 - x + 1/2^2 - 1/2^2) + 1
y = -1/2(x^2 -x + 1/4 - 1/4) + 1
y = -1/2(x^2 -x + 1/4) - 1/2(-1/4) + 1
y = -1/2(x - 1/2)(x - 1/2) + 1/8 + 1(8)/8
y = -1/2(x - 1/2)^2 + 9/8
BUT use factored form
we know that
x = -1 and x = 2
x+1 = 0 and x - 2 = 0
SO our equation in factored form is
y = a(x+1)(x-2)
and sub vertex in for x and y to solve for a
1 = a(1+1)(1-2)
1 = a(2)(-1)
-2a = 1
a = -1/2
THEREFORE our equation is y = -1/2(x+1)(x-2)
y = -1/2(x^2 +x -2x -2)
y = -1/2(x^2 -x - 2)
y = -1/2x^2 + 1/2x + 1
which if you want to get it into vertex form
y = -1/2(x^2 - x) + 1
y = -1/2(x^2 - x + 1/2^2 - 1/2^2) + 1
y = -1/2(x^2 -x + 1/4 - 1/4) + 1
y = -1/2(x^2 -x + 1/4) - 1/2(-1/4) + 1
y = -1/2(x - 1/2)(x - 1/2) + 1/8 + 1(8)/8
y = -1/2(x - 1/2)^2 + 9/8