+0  
 
0
154
1
avatar+164 

Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0).  What is (a, b, c)?

xXxTenTacion  Jul 16, 2018

Best Answer 

 #1
avatar+94201 
+2

Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0).  What is (a, b, c)?

 

roots are    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

 

axis of symmetry is   

\(x= \frac{-b}{2a}\\ \frac{-b}{2a}=3\\ -b=6a\\ b=-6a\\ \)

 

\(y=ax^2+bx+c\\ y=ax^2-6ax+c\\ sub \;\;in \;\;(1,0)\\ 0=a-6a+c\\ 0=c-5a\\ c=5a \)       

 

 

\(y=ax^2-6ax+c\\ y=ax^2-6ax+5a\\ sub\;in\; (3,2)\\ 2=9a-18a+5a\\ 2=-4a\\ a=-\frac{1}{2}\\ \)

 

a=-0.5

c=-2.5

b=3

 

I do not have time to check so you had better do it :)

Melody  Jul 16, 2018
 #1
avatar+94201 
+2
Best Answer

Suppose that we have an equation y=ax^2+bx+c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0).  What is (a, b, c)?

 

roots are    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

 

axis of symmetry is   

\(x= \frac{-b}{2a}\\ \frac{-b}{2a}=3\\ -b=6a\\ b=-6a\\ \)

 

\(y=ax^2+bx+c\\ y=ax^2-6ax+c\\ sub \;\;in \;\;(1,0)\\ 0=a-6a+c\\ 0=c-5a\\ c=5a \)       

 

 

\(y=ax^2-6ax+c\\ y=ax^2-6ax+5a\\ sub\;in\; (3,2)\\ 2=9a-18a+5a\\ 2=-4a\\ a=-\frac{1}{2}\\ \)

 

a=-0.5

c=-2.5

b=3

 

I do not have time to check so you had better do it :)

Melody  Jul 16, 2018

3 Online Users

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.