# Math Pre-Cal

When I am giving the point vertex of (1,-1) and another point of (2,2) and i need to put the answere in the form f(x)=ax^2+bx+c how would I go about doing that?

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1. Use the vertex form first

y = a(x-h)^2 + k, where (h, k) is the vertex, so ...

vertex is (1,-1)
so f(x) = a(x-1)^1 - 1
but (2,2) lies on it, so
2 = a(2-1)^2 - 1
2 = a(1) - 1
3 = a

so you have f(x) = 3(x-1)^2 - 1
now expand this ...
f(x) = 3(x^2 - 2x + 1) - 1
= 3x^2 - 6x + 3 - 1
= 3x^2 - 6x + 2

(Notice, both points satisfy the equation.)

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posted by Reiny
2. Ok I understand now on that so far. How do you find the value of a quadratic function?

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posted by Holly
3. to know the "value" of the quadratic function, you must know the value of x
simply plug in your given value of x and work it out.

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posted by Reiny
4. what is the value of f(x)=2x^2-12x-6

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posted by Holly
5. It ask for the Min or Max value?

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posted by Holly
6. The answer I got was 24, and it was wrong

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posted by Holly
7. If you do not know calculus you must find the vertex of that parabola.

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posted by Damon
8. how would i do that

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posted by Holly
9. i really need the help i am so lost

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posted by Holly
10. If you do know calculus, find the derivative and set it equal to zero (the slope is 0 at a max or min)

4 x -12 = 0
so
x = 3
then
y = 2x^2-12x-6 = 2(9) -12(3) -6
=18 - 36 - 6
= -24
that vertex is a minimum because y gets big as x gets big

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posted by Damon
11. to do it with algebra, complete the square

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posted by Damon
12. 2 x^2 -12 x - 6 = y
x^2 - 6 x - 3 = y/2

x^2 - 6x = y/2 + 3

x^2 - 6 x + 9 = y/2 + 12

(x-3)^2 = (1/2)(y + 24)
vertex at x = 3 and y = -24

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posted by Damon
13. you can also make use of the fact that the vertex is midway between the roots.

Recall that the roots are at x = -b/2a ±(some junk)

That means that the vertex is at x = -b/2a. Plug that in to get the y-value:

y = 2x^2-12x-6

the vertex is at x = 12/4 = 3
y(3) = 2*9-12*3-6 = -24

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posted by Steve

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