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# Algebra

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From the quadratic equations from the following:

Problem 1:

If

f(2)=f(4)=0, f(3)=−2.

Problem 2:

If

Zeroes at 1± sqrt(2),  y-intercept at –4.

Guest Feb 26, 2017
#1
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Problem 1:

If

f(2)=f(4)=0, f(3)=−2

x = 2  and x = 4 are roots    and the vertex  lies at  (3, -2)......so we know that

0  = a(2 - 3)^2  - 2   simplify

0 = a (-1)^2  - 2

0 =  a - 2   →   a = 2

So...our function is f(x)  = 2(x - 3)^2 - 2

Check   that x = 4 produces a zero

2(4 - 3)^2 - 2    =   2(1)^2  - 2   =   2 - 2  = 0

Problem 2:

If

Zeroes at 1± sqrt(2),  y-intercept at –4

The  x value of the vertex will be the average of the sum of the roots =

[  1 + sqrt (2) +  1 - sqrt(2) ] / 2    = 2 / 2   = 1

And the x coordinate of the vertex  =  -B/ [ 2A]   implies that

-B/ [2A]  = 1   which implies that

-B  =  2A   which implies that

B = -2A

So  we have the following :

y  =  Ax^2  + Bx + C     and substituting for B

y = Ax^2  - (2A)x + C

And since the y intercept is at - 4 , then

-4  = A(0)^  -2A(0)  + C

-4  =  0 + 0 + C

-4 = C

So we have that

y = Ax^2 -2Ax - 4

And we know that

0  =  A(1 + sqrt(2))^2  - 2A(1 + sqrt(2))   - 4

4 = A ( 1 + 2sqrt(2) + 2)  - 2A  - 2Asqrt(2)

4 =  A + 2Asqrt(2) + 2A - 2A - 2Asqrt(2)

4 = A       ....    and  B  = -2A =  -8

So  we have

y = 4x^2 - 8x - 4

Here's a graph that confirms this : https://www.desmos.com/calculator/ba8scvgskt

CPhill  Feb 26, 2017