Answer:
- vertex form: y=a(x-h)^2+k ⇒ (a, h, k) = (-1/4, -2, 4)
- factored form: y=a(x+6)(x-2) ⇒ a = -1/4
- standard form: y=ax^2+bx+c ⇒ (a, b, c) = (-1/4, -1, 3)
Explanation:
We don't know how you did the first way, so the only way we can make sure another way is different is to do it two ways.
1) First, we'll make use of the vertex-form equation. The given zeros are at -6 and +2, so the line of symmetry is x = (-6+2)/2 = -2. Then vertex form of the equation will look like ...
y = a(x +2)^2 +k
We have two unknown values, so we need to make use of two points to find them.
Using (2, 0):
0 = a(2 +2)^2 +k
0 = 16a +k
Using (-4, 3):
3 = a(-4+2)^2 +k
3 = 4a +k
Subtracting the second equation from the first, we get
0 -3 = (16a +k) -(4a +k)
-3 = 12a
a = -1/4
Substituting into the first of our equations, ...
0 = 16(-1/4) + k = -4 +k
k = 4
So, the equation is ...
y = (-1/4)(x +2)^2 +4
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2) Using factored form, we can write the equation as ...
y = a(x +6)(x -2) . . . . for zeros x=-6 and x=2
We can find 'a' by using the third point:
3 = a(-4+6)(-4-2) = a(2)(-6) = -12a
a = 3/-12 = -1/4
So, the equation is ...
y = (-1/4)(x +6)(x -2)
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3) You can also make use of the standard form equation. This gives rise to three equations in the three coefficients.
y = ax^2 +bx +c
0 = a(-6)^2 +b(-6) +c = 36a -6b +c
3 = a(-4)^2 +b(-4) +c = 16a -4b +c
0 = a(2)^2 +b(2) +c = 4a +2b +c
These can be solved by your favorite method to give ...
a = -1/4, b = -1, c = 3
So, the equation is ...
y = (-1/4)x^2 -x +3