+0  
 
0
259
1
avatar

The graph of \(y=ax^2 + bx + c\) is a parabola with vertical axis of symmetry. The vertex of this parabola is \((2,3)\) and the parabola contains the point \((4,4)\). Find the value of \(y\) when \(x=6\).

Guest Jan 14, 2018
 #1
avatar+91027 
+1

y = ax^2 + bx  +  c

 

The x coordinate of the vertex  =  -b / [2a]  =  2  ⇒  b = - 4a    

 

We know that

 

a(2)^2 + b(2)  +  c  =  3  ⇒  4a  + 2b  +  c  =  3  ⇒  4a -  8a  + c  =  3  ⇒  -4a + c = 3    (1)

a(4)^2 + b(4)  +  c  =  4  ⇒ 16a  + 4b  + c  = 4  ⇒  16a - 16a + c  = 4  ⇒  c  = 4   (2)

 

Subbing (2)  into (1)  we have that   

-4a  +  4 =  3

-4a  =  -1

a  = -1/-4  =  1/4

 

And     b  =  -4(1/4)  =  -1

 

So......the function is 

 

y  = (1/4)x^2  -  x  +  4

 

So....when  x  = 6 

y  =  (1/4)6^2 - (6)  +  4  =  9  - 6  +  4   =   7

 

 

cool cool cool

CPhill  Jan 14, 2018

38 Online Users

avatar
avatar

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.