Question

# Completing the square is also possible when the coefficient of the x² term is not 1. Work your way through this problem to see how. y = 3x² - 18x + 1

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Completing the square is also possible when the coefficient of the x² term is not 1. Work your way through this problem to see how. $$\displaystyle{y}={3}{x}²-{18}{x}+{1}$$
$$\displaystyle{y}-{1}={3}{\left({x}²-{6}{x}\right)}$$ a. What number needs to be added to both sides of the equation so that the expression inside the parentheses is a perfect square? Hint: 9 is the wrong answer.
b. Finish completing the square and find the vertex of the parabola.

2021-05-23
a.For $$\displaystyle{x}^{{2}}−{6}{x}$$ to be a perfect square trinomial, you must add $$\displaystyle{\left(\frac{{b}}{{2}}\right)}{2}={\left(−\frac{{6}}{{2}}\right)}^{{2}}={9}$$. This leads us to:
$$\displaystyle{y}−{1}={3}{\left({x}^{{2}}−{6}{x}+{9}\right)}$$
To balance what we added to the right side, consider the expanded form:
$$\displaystyle{y}−{1}={3}{x}^{{2}}−{18}{x}+{27}$$
Hence, you must add 27 to the left side:
$$\displaystyle{y}−{1}+{27}={3}{x}^{{2}}−{18}{x}+{27}$$
$$\displaystyle{y}+{26}={3}{\left({x}^{{2}}−{6}{x}+{9}\right)}$$
b.Write the parabola in vertex form y−k=a(x−h)2 where (h,k) is the vertex. Here, we have:
y+26=3(x−3)2
or
y−(−26)=3(x−3)2
So, the vertex is at (3, -26).