+0  
 
+5
148
1
avatar

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
avatar+87309 
0

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

 

 

cool cool cool

CPhill  Feb 26, 2017

7 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.