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i need help please 

 
Guest Jul 12, 2018

Best Answer 

 #1
avatar+19636 
+1

i need help please

 

\(\begin{array}{|lrcll|} \hline & y &=& x^2+bx+c \\ \hline (0,14): & -14 &=& 0^2+b\cdot 0 + c \\ & -14 & = & c \\ & \mathbf{c} & \mathbf{=} & \mathbf{-14} \\ \hline (2,0): & 0 &=& 2^2+b\cdot 2 + c \quad & | \quad c = -14 \\ & 0 &=& 2^2+b\cdot 2 -14 \\ & 0 &=& 4+2b -14 \\ & 2b &=& 14-4 \\ & 2b &=& 10 \\ & \mathbf{b} & \mathbf{=} & \mathbf{5} \\ \hline & \mathbf{y} & \mathbf{=} & \mathbf{x^2+5x-14} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x^2+5x-14 &=& 0 \\ x &=& \dfrac{-5\pm \sqrt{25-4\cdot (-14) } }{2} \\ &=& \dfrac{-5\pm \sqrt{81} }{2} \\ &=& \dfrac{-5\pm 9 }{2} \\\\ x_1 &=& \dfrac{-5+ 9 }{2} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2} \\\\ x_2 &=& \dfrac{-5- 9 }{2} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-7} \\ \hline \end{array}\)

 

Point P at (-7,0)

\(\begin{array}{|rcll|} \hline x_{\text{turning point}} &=& \dfrac{x_1+x_2}{2} \\\\ &=& \dfrac{2-7}{2} \\\\ &=& \dfrac{-5}{2} \\\\ &\mathbf{=} & \mathbf{-2.5} \\ \\ \hline y_{\text{turning point}} &=& (-2.5)^2 +5\cdot (-2.5) -14 \\ &=& 6.25 -12.5 -14 \\ &\mathbf{=} & \mathbf{-20.25} \\ \hline \end{array}\)

 

The x-coordinate of the turning point is -2.5

 

 

laugh

 
heureka  Jul 12, 2018
 #1
avatar+19636 
+1
Best Answer

i need help please

 

\(\begin{array}{|lrcll|} \hline & y &=& x^2+bx+c \\ \hline (0,14): & -14 &=& 0^2+b\cdot 0 + c \\ & -14 & = & c \\ & \mathbf{c} & \mathbf{=} & \mathbf{-14} \\ \hline (2,0): & 0 &=& 2^2+b\cdot 2 + c \quad & | \quad c = -14 \\ & 0 &=& 2^2+b\cdot 2 -14 \\ & 0 &=& 4+2b -14 \\ & 2b &=& 14-4 \\ & 2b &=& 10 \\ & \mathbf{b} & \mathbf{=} & \mathbf{5} \\ \hline & \mathbf{y} & \mathbf{=} & \mathbf{x^2+5x-14} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x^2+5x-14 &=& 0 \\ x &=& \dfrac{-5\pm \sqrt{25-4\cdot (-14) } }{2} \\ &=& \dfrac{-5\pm \sqrt{81} }{2} \\ &=& \dfrac{-5\pm 9 }{2} \\\\ x_1 &=& \dfrac{-5+ 9 }{2} \\ \mathbf{x_1} & \mathbf{=} & \mathbf{2} \\\\ x_2 &=& \dfrac{-5- 9 }{2} \\ \mathbf{x_2} & \mathbf{=} & \mathbf{-7} \\ \hline \end{array}\)

 

Point P at (-7,0)

\(\begin{array}{|rcll|} \hline x_{\text{turning point}} &=& \dfrac{x_1+x_2}{2} \\\\ &=& \dfrac{2-7}{2} \\\\ &=& \dfrac{-5}{2} \\\\ &\mathbf{=} & \mathbf{-2.5} \\ \\ \hline y_{\text{turning point}} &=& (-2.5)^2 +5\cdot (-2.5) -14 \\ &=& 6.25 -12.5 -14 \\ &\mathbf{=} & \mathbf{-20.25} \\ \hline \end{array}\)

 

The x-coordinate of the turning point is -2.5

 

 

laugh

 
heureka  Jul 12, 2018

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