+0  
 
+5
77
2
avatar

what is the vertex of this equation 3x^2-11x-4=y

Guest Mar 6, 2017

Best Answer 

 #1
avatar+75352 
+5

3x^2-11x-4=y

 

We have the form

 

Ax^2 + Bx + C   = y

 

The x coordinate of the vertex is given by :  -B/ (2A)  = - (-11) / (2*3)  = 11/6

 

To find the y coordinate.....put this x value back into the function....and we have

 

3(11/6)^2 - 11(11/6) - 4

 

3(121/36) - 121/6 - 4

 

121/12 - 121/6 - 4

 

121/12 - 242/12 - 48/12  =

 

-169/12

 

So....the vertex is ( 11/6, -169/12 )

 

 

cool cool cool

CPhill  Mar 6, 2017
Sort: 

2+0 Answers

 #1
avatar+75352 
+5
Best Answer

3x^2-11x-4=y

 

We have the form

 

Ax^2 + Bx + C   = y

 

The x coordinate of the vertex is given by :  -B/ (2A)  = - (-11) / (2*3)  = 11/6

 

To find the y coordinate.....put this x value back into the function....and we have

 

3(11/6)^2 - 11(11/6) - 4

 

3(121/36) - 121/6 - 4

 

121/12 - 121/6 - 4

 

121/12 - 242/12 - 48/12  =

 

-169/12

 

So....the vertex is ( 11/6, -169/12 )

 

 

cool cool cool

CPhill  Mar 6, 2017
 #2
avatar+18389 
+5

what is the vertex of this equation 3x^2-11x-4=y

 

Formula:

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\\\ x_v &=& -\frac{b}{2a} \\ y_v &=& c - \frac{b^2}{4a} \\ \hline \end{array}\)

 

Vertex:

\(\begin{array}{|rcll|} \hline y &=& 3x^2-11x-4 \quad & | \qquad a = 3 \qquad b=-11 \qquad c = -4 \\\\ x_v &=& -\frac{(-11)}{2\cdot 3} \\ &=& \frac{11}{6} \\ \mathbf{x_v} &\mathbf{=}& \mathbf{1.8\bar{3}} \\\\ y_v &=& -4 - \frac{(-11)^2}{4\cdot 3} \\ &=& -4 - \frac{121}{12} \\ &=& -\frac{169}{12} \\ \mathbf{y_v} &\mathbf{=}& \mathbf{-14.08\bar{3}} \\ \hline \end{array}\)

 

laugh

heureka  Mar 6, 2017

5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details