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FreeTrader

  • 3 years ago

Problem Set 1 Question 1A-1 b) By completing the square, use translation and change of scale to sketch: y = 3x^2 + 6x +2 Here is the method I attempted, but with a different result than shown in the Solutions. Where am I going wrong? y = 3x^2 + 6x +2 y - 2 = 3x^2 + 6x [move 2 to the left side] y - 2 = 3(x^2 +2x) [factor out 3] y - 2 +1 = 3(x^2 +2x +1) [complete the square] y - 1 = 3(x +1)^2 [rearrange toward slope - intercept form] y = 3(x+1)^2 +1 [add one to both sides] Now, I have it in slope-intercept form but my answer is different from the Solution. Help?

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  1. LagrangeSon678
    • 3 years ago
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    you could have factored out a 3 from the begining

  2. FreeTrader
    • 3 years ago
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    The answer for the slope-intercept form given by the Solutions set is different from mine, it is \[y=3(x+1)^{2}-1\] The last number I got was +1, but the key says -1. Am I doing something wrong?

  3. FreeTrader
    • 3 years ago
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    I now see that in step 4, when I completed the square, I should add 3 to the left side due to the 3 distribution on the 1 on the right side. This results in a solution that matches the answer key, and is equivalent to the original equation when substituting in various values for x. Thanks to Throol, who addressed this in another forum.

  4. FreeTrader
    • 3 years ago
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    LagrangeSon678, you are right. Dividing out the 3 on the right side also would work, and would eliminate the need to distribute the three over the added 1 on the right. As follows: 1) Original Equation: \[y=3x ^{2}+6x+2\] 2) Divide out 3 from Right Side: \[y/3 = x^2 +2x + 2/3\] 3) Subtract 2/3 from both sides: \[y/3 -2/3 = x^2 +2x\] 4) Complete the Square: \[y/3 -2/3 + 1 = x^2 +2x +1\] 5) Get Common Denominator on Left Side: \[y/3 -2/3 + 3/3 = x^2 +2x +1\] 6) Simplify Left Side: \[ y/3 + 1/3 = x^2 +2x +1\] 7) Factor Right Side: \[ y/3 + 1/3 = (x+1)^2\] 8) Multiply by 3 on both sides: \[y +1 = 3(x+1)^2\] 9) Subtract 1 from both sides: \[y = 3(x+1)^2 - 1\] And now we have the equation in slope-intercept form, and it can be sketched readily.

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