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# what is the vertex?

+5
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what is the vertex of this equation 3x^2-11x-4=y

Guest Mar 6, 2017

#1
+87309
+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 )

CPhill  Mar 6, 2017
#1
+87309
+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 )

CPhill  Mar 6, 2017
#2
+19653
+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:

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

heureka  Mar 6, 2017