The equation of a curve is $$\(y=2 x^{2}+k x+k-1\)$$, where $$\(k\)$$ is a constant. It is now given that $$\(k=2\)$$. Express the equation of the curve in the form $$\(y=2(x+a)^{2}+b\)$$, where $$\(a\)$$ and $$\(b\)$$ are constants, and hence state the coordinates of the vertex of the curve. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s20_qp_12 Year：2020 Question No：6(b)
Answer:
\[
2 x^{2}+2 x+1=2\left(x+\frac{1}{2}\right)^{2}+1-\frac{1}{2}
\]
\[
a=\frac{1}{2}, b=\frac{1}{2}
\]
\(\operatorname{vertex}\left(-\frac{1}{2}, \frac{1}{2}\right)\)
(FT on \(a\) and \(b\) values)
2 x^{2}+2 x+1=2\left(x+\frac{1}{2}\right)^{2}+1-\frac{1}{2}
\]
\[
a=\frac{1}{2}, b=\frac{1}{2}
\]
\(\operatorname{vertex}\left(-\frac{1}{2}, \frac{1}{2}\right)\)
(FT on \(a\) and \(b\) values)
Knowledge points:
1.1.1 carry out the process of completing the square for a quadratic polynomial and use a completed square form
Solution:
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