Explanation Given: Consider the exercises of system of equations, { y = − x + 2 y = x 2 − 4 { y = − x + 2 y = y = x 2 − 4 x + 4 And, { y = x 2 − 4 x − 10 y = − x 2 − 2 x + 14 { y = 6 x − 1 y = 2 x − 1 And, { y = ( − x 2 4 + 5 ) 1 2 y = ( − x 2 + 3 ) Proof: For first set of equation, Find the graph and solution of system of equation y = − x + 2 and y = x 2 − 4 by using TI-83 are as follows: Step 1: Open TI-83 and then press Y button. Step 2: Enter the first equation Y 1 = − x + 2 . Step 3: Press DOWN key and enter the second equation Y 2 = x 2 − 4 . Step 4: After entering the values, then press GRAPH key. Now, find the solution of equations by using intersection features of Ti-83. Press TRACE key and trace the graph, When the trace sign in right side of origin, then, Step 1: Press 2 nd and TRACE. Step 2: Select number 5 th option of (Intersects). Step 3: Press ENTER for intersections of first function. Step 4: Again, press ENTER for intersections of the second function. Step 5: Again, press ENTER for guess. Step 6: Again, press ENTER for intersection. So, the intersection point is obtained ( 2 , 0 ) . When the trace sign in left side of origin, then, The solution of equations by using intersection features, Step 1: Press (2 nd ) and TRACE. Step 2: Select number 5 th option of (Intersect). Step 3: Press ENTER for intersections of first function. Step 4: Again, press ENTER for intersections of the second function. Step 5: Again, press ENTER for guess. Step 6. Again, press ENTER for intersection. So, the intersection point obtained is ( − 3 , 5 ) . Hence, the intersection points are ( − 3 , 5 ) and ( 2 , 0 ) . For the second system of equations, find the graph and solution of system of equations y = − x + 2 and y = x 2 − 4 x + 4 by using TI-83. Draw the graph by using TI-83 by the following steps, Step 1: Open TI-83 and then press Y button. Step 2: Enter first equation Y 1 = − x + 2 . Step 3: Press DOWN key and enter the second equation Y 2 = x 2 − 4 x + 4 . Step 4: After entering the values, then press GRAPH key. Now, find the solution of equations by using intersection features of Ti-83. Press TRACE key and trace the graph, When the trace sign in right side of origin, then, Step 1: Press (2 nd ) and TRACE. Step 2: Select number 5 th option of (Intersects). Step 3: Press ENTER for intersections of fist function. Step 4: Again, press ENTER for intersections of the second function. Step 5: Again, press ENTER for guess. Step 6: Again, press ENTER for intersection. So, the intersection point obtained is ( 2 , 0 ) . When the trace sign in left side of origin, then, The solution of equations by using intersection features. Step 1: Press (2 nd ) and TRACE. Step 2: Select number 5 th option of (Intersects). Step 3: Press ENTER for intersections of fist function. Step 4: Again, press ENTER for intersections of the second function. Step 5: Again, press ENTER for guess. Step 6: Again, press ENTER for intersection. So, the intersection point obtained is ( 1 , 1 ) . Hence, the intersection points are ( 2 , 0 , ) and ( 1 , 1 )

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ISBN: 9781323142707

Author: Blitzer

Publisher: Pearson Custom Publishing

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Chapter 8.4, Problem 67E

To determine

To prove: The solution by using graphical utility from the five pair of equations,

{y=−x+2y=x2−4{y=−x+2y=y=x2−4x+4

And,

{y=x2−4x−10y=−x2−2x+14{y=6x−1y=2x−1

And,

{y=(−x24+5)12y=(−x2+3)

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Chapter 8.4, Problem 66E

Chapter 8.4, Problem 68E

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Chapter 8 Solutions

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To prove: The solution by using graphical utility from the five pair of equations, { y = − x + 2 y = x 2 − 4 { y = − x + 2 y = y = x 2 − 4 x + 4 And, { y = x 2 − 4 x − 10 y = − x 2 − 2 x + 14 { y = 6 x − 1 y = 2 x − 1 And, { y = ( − x 2 4 + 5 ) 1 2 y = ( − x 2 + 3 )

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To prove: The solution by using graphical utility from the five pair of equations, {y=−x+2y=x2−4{y=−x+2y=y=x2−4x+4 And, {y=x2−4x−10y=−x2−2x+14{y=6x−1y=2x−1 And, {y=(−x24+5)12y=(−x2+3)

Want to see the full answer?

Chapter 8 Solutions

To prove: The solution by using graphical utility from the five pair of equations,

{y=−x+2y=x2−4{y=−x+2y=y=x2−4x+4

And,

{y=x2−4x−10y=−x2−2x+14{y=6x−1y=2x−1

And,

{y=(−x24+5)12y=(−x2+3)