• Dec 28th 2005, 07:19 AM
danielle_54321
The question is:
A parabola has zeros -6 and 2 and y-intercept -9.
a) Write its equation in factored form, expanded form and vertex form.
b)Use the expanded form and the quadratic formula to verify that the zeros are -6 and 2.

First, to write the equation in factored form I substituted y=-9, x=0 to find a.
y=a(x-s)(x-t)
-9=a(0+6)(0-2)
-9=a(-6)(2)
-9=-12a
3/4=a

Which therefore made the equation in factored form:
y=3/4(x+6)(x-2)

And it was therefore in expanded form:

y=3/4x^2+3x-9

However, in checking the zeros of this equation with the quadratic formula,
I could not get -6 and 2 for the zeros.

x=-bħsquare root of b^+4ac/2a
x=-3ħsquare root of 3x3+4(3/4)(-9)/2a
x=-3ħsquare root of 9-27/2a
x=-3ħsquare root of -18 /2a

Already I know that this is wrong.
I must email this to my teacher before the end of the week.
• Dec 28th 2005, 08:18 AM
MathGuru
You solved for the equation correctly but your quadratic equation is incorrect.

Eq: (3/4)x^2+3x-9

a=.75
b=3
c=-9

-bħsqrt(b^2 - 4*a*c)
---------------------
.............2*a

-3ħsqrt(3^2 - 4*.75*-9)
---------------------
.............2*.75

-3ħsqrt(9 + 27)
---------------
.........1.5

-3ħ6
-----
.1.5

-3+6
-----=2
.1.5

-3-6
-----=-6
.1.5
• Dec 28th 2005, 08:29 AM
CaptainBlack
Quote:

Originally Posted by danielle_54321
The question is:
A parabola has zeros -6 and 2 and y-intercept -9.
a) Write its equation in factored form, expanded form and vertex form.
b)Use the expanded form and the quadratic formula to verify that the zeros are -6 and 2.

First, to write the equation in factored form I substituted y=-9, x=0 to find a.
y=a(x-s)(x-t)
-9=a(0+6)(0-2)
-9=a(-6)(2)
-9=-12a
3/4=a

Which therefore made the equation in factored form:
y=3/4(x+6)(x-2)

And it was therefore in expanded form:

y=3/4x^2+3x-9

However, in checking the zeros of this equation with the quadratic formula,
I could not get -6 and 2 for the zeros.

x=-bħsquare root of b^+4ac/2a
x=-3ħsquare root of 3x3+4(3/4)(-9)/2a
x=-3ħsquare root of 9-27/2a
x=-3ħsquare root of -18 /2a

Already I know that this is wrong.
I must email this to my teacher before the end of the week.

The quadratic formula for the roots is:

$\displaystyle x=\frac{-b\pm {\sqrt{b^2-4ac}}}{2a}$

substituting $\displaystyle a=3/4,\ b=3,\ c=-9$ will give the required roots.

RonL
• Dec 29th 2005, 02:35 AM
ticbol
Quote:

Originally Posted by danielle_54321
The question is:
A parabola has zeros -6 and 2 and y-intercept -9.
a) Write its equation in factored form, expanded form and vertex form.
b)Use the expanded form and the quadratic formula to verify that the zeros are -6 and 2.

First, to write the equation in factored form I substituted y=-9, x=0 to find a.
y=a(x-s)(x-t)
-9=a(0+6)(0-2)
-9=a(-6)(2)
-9=-12a
3/4=a

Which therefore made the equation in factored form:
y=3/4(x+6)(x-2)

And it was therefore in expanded form:

y=3/4x^2+3x-9

Umm, this is the first time I saw how a parabola is treated this way. It is unusual for me. I saw this yesterday and it intrigued me. Then a while ago, on my way home, I saw reason behind it. Nice.

y = Ax^2 +Bx +C
is a parabola. It being a vertical parabola, it
is also a function of x, or y=f(x), because an imaginary vertical line will cross this parabola at only one point per position of the vertical line along the x-axis.
The zeroes at -6 and 2 mean they are zeroes of the function. And that was the reason why y = a(x+6)(x-2), or why it is a factored form of the vertical parabola. Zeroes---factors---polynomial. Umm.

I am used to [y = a(x-h)^2 +k] for a vertical parabola.
[ y = a(x-s)(x-t), whatever s and t mean, is new to me, as I have said. ]
where (h,k) is the vertex.
Or, it is more like: (y-k) = a(x-h)^2 -----------(i)
I guess that is what you call above as the vertex form of the parabola.

So you got the factored form and the expanded form. What about the vertex form?

Let us do it.

As you can see from the righthand side of Eq.(i), the (x-h)^2 is a perfect square, or (x-h) is multiplied by itself. That means we just "complete the square" of the x-terms in any of the factored or expanded forms to get the vertex form.
Let's use, say, the factored form:
y = (3/4)(x+6)(x-2)
To "complete the square", we make sure first that the coefficient of the x^2 term is 1 only,
y = (3/4)[x^2 +4x -12]
y = (3/4)[(x^2 +4x) -12]
y = (3/4)[(x^2 +4x +(4/2)^2 -(4/2)^2 -12]
y = (3/4)[(x+2)^2 -4 -12]
y = (3/4)[(x+2)^2 -16]
y = (3/4)(x+2)^2 -(3/4)(16)
y = (3/4)(x+2)^2 -12 ------------------vertex form, or,
(y +12) = (3/4)(x+2)^2 ------------vertex form.

That means the vertex is (-2,-12). --------***

-----------------------
If you did not know the factored form way, by using the vertex form you can also get the equation of the parabola:
(y-k) = a(x-h)^2
at point (-6,0)....(0 -k) = a(-6 -h)^2 ----(1)
at point (2,0).....(0 -k) = a(2 -h)^2 --------(2)
at point (0,-9)....(-9 -k) = a(0 -h)^2 -----------(3)
3 equations, 3 unknowns, solvable.

Take care only when you equate the (-6 -h) of (1) to the (2 -h) of (2).
(-6 -h) = (2 -h)
Cannot be.
But (-6 -h)^2 is also [-(6+h)]^2 = (6+h)^2, so,
a(-6 -h)^2 = a(2 -h)^2
(6+h)^2 = (2-h)^2
6+h = 2-h
h +h = 2 -6
2h = -4
h = -2 -----------the x-coordinate of the vertex.

Etc..
• Dec 29th 2005, 05:33 AM
danielle_54321
Thank-you so much to all of you; your advise was very insightful. :)