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Please and Thanks (Help!)angry

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The diagram shows a calculator screen on which the parabolas y=1/4(x-3)(x-8) and y=1/2(x+1)(x-3) have been graphed. The window setting consists of two inequalities, A is less than or equal to X is less than or equal to B and C is less than or equal to Y is less than or equal to D. What are the values of a, b, c, and d? Explain your reasoning, please!!

SydSu22  Dec 7, 2018
edited by SydSu22  Dec 7, 2018

Best Answer 

 #1
avatar+20683 
+4

th 5 prob 5

The diagram shows a calculator screen on which the parabolas y=1/4(x-3)(x-8) and y=1/2(x+1)(x-3) have been graphed. The window setting consists of two inequalities, A is less than or equal to X is less than or equal to B and C is less than or equal to Y is less than or equal to D. What are the values of a, b, c, and d? Explain your reasoning, please!!

 

\(\mathbf{y=\dfrac14(x-3)(x-8)}\\ \begin{array}{|lrcll|} \hline x =8 ~\Rightarrow~ y = 0 ~\Rightarrow~ b=8 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline x =0 ~\Rightarrow~ &y&=&\dfrac14(-3)(-8) \\ & y&=&\dfrac14\cdot 3 \cdot 8 \\ & y &=& 6 ~\Rightarrow~ d=6 \\ \hline \end{array}\)

 

\(\mathbf{y=\dfrac12(x+1)(x-3)}\\ \begin{array}{|lrcll|} \hline x_{\text{vertex}} &=& \dfrac{-1+3}{2} \\ &=& 1 \\\\ y_{\text{vertex}} &=&\dfrac12(1+1)(1-3) \\ &=&-2 ~\Rightarrow~ c=-2 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline y = 6 ~\Rightarrow~ &6& =&\dfrac12(x+1)(x-3) \\ &12&=& (x+1)(x-3) \\ & x^2-2x-3 &=& 12 \\ & x^2-2x-15 &=& 0 \\ & x &=& \dfrac{2\pm \sqrt{4-4\cdot (-15)}}{2} \\ & x &=& \dfrac{2\pm \sqrt{64}}{2} \\ & x &=& \dfrac{2\pm 8}{2} \\\\ & x &=& \dfrac{2+ 8}{2} \\ & x &=& 5 \\\\ & x &=& \dfrac{2- 8}{2} \\ & x &=& -3 ~\Rightarrow~ a=-3 \\ \hline \end{array}\)

 

\(-3 \le x \le 8 \\ -2 \le y \le 6\)

 

laugh

heureka  Dec 7, 2018
 #1
avatar+20683 
+4
Best Answer

th 5 prob 5

The diagram shows a calculator screen on which the parabolas y=1/4(x-3)(x-8) and y=1/2(x+1)(x-3) have been graphed. The window setting consists of two inequalities, A is less than or equal to X is less than or equal to B and C is less than or equal to Y is less than or equal to D. What are the values of a, b, c, and d? Explain your reasoning, please!!

 

\(\mathbf{y=\dfrac14(x-3)(x-8)}\\ \begin{array}{|lrcll|} \hline x =8 ~\Rightarrow~ y = 0 ~\Rightarrow~ b=8 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline x =0 ~\Rightarrow~ &y&=&\dfrac14(-3)(-8) \\ & y&=&\dfrac14\cdot 3 \cdot 8 \\ & y &=& 6 ~\Rightarrow~ d=6 \\ \hline \end{array}\)

 

\(\mathbf{y=\dfrac12(x+1)(x-3)}\\ \begin{array}{|lrcll|} \hline x_{\text{vertex}} &=& \dfrac{-1+3}{2} \\ &=& 1 \\\\ y_{\text{vertex}} &=&\dfrac12(1+1)(1-3) \\ &=&-2 ~\Rightarrow~ c=-2 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline y = 6 ~\Rightarrow~ &6& =&\dfrac12(x+1)(x-3) \\ &12&=& (x+1)(x-3) \\ & x^2-2x-3 &=& 12 \\ & x^2-2x-15 &=& 0 \\ & x &=& \dfrac{2\pm \sqrt{4-4\cdot (-15)}}{2} \\ & x &=& \dfrac{2\pm \sqrt{64}}{2} \\ & x &=& \dfrac{2\pm 8}{2} \\\\ & x &=& \dfrac{2+ 8}{2} \\ & x &=& 5 \\\\ & x &=& \dfrac{2- 8}{2} \\ & x &=& -3 ~\Rightarrow~ a=-3 \\ \hline \end{array}\)

 

\(-3 \le x \le 8 \\ -2 \le y \le 6\)

 

laugh

heureka  Dec 7, 2018
 #2
avatar+105 
+1

Thank you so much you are a life saver!!

SydSu22  Dec 7, 2018

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