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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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?

OCW Scholar - Single Variable Calculus
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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In your fourth step, you've added 1 to the left side, but 3 to the right (3(+1), so the quantities are no longer equal. Try adding 3 to both sides, and you should get what you're looking for :)
\[y-2=3(x ^{2}+2x)\] Next you'll add 3 to both sides: \[y-2+3=3(x ^{2}+2x)+3\] Factor out the 3, \[y+1=3(x ^{2}+2x+1)\] and take it from there.
Thank you, Throol. I see that your solution is equivalent to the original equation.

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