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# th 5 prob 5

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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!!

Dec 7, 2018
edited by SydSu22  Dec 7, 2018

#1
+22286
+11

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!!

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

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

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

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

$$-3 \le x \le 8 \\ -2 \le y \le 6$$

Dec 7, 2018

#1
+22286
+11

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!!

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

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

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

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

$$-3 \le x \le 8 \\ -2 \le y \le 6$$

heureka Dec 7, 2018
#2
+250
+1

Thank you so much you are a life saver!!

SydSu22  Dec 7, 2018