What is \(y=2(x+3)^{2}+1\) in standard form?
Answer:
\(y=2(x+3)^{2}+1\) in standard form is \(y=2 x^{2}+12 x+19\).
Explanation:
\(y=2(x+3)^{2}+1\) is a quadratic equation in vertex form. It can be converted into standard form by doing the following:
Remove the parentheses and exponent by simplifying \((x+3)^{2}\) which is a sum of squares.
\(a^{2}+b^{2}=a^{2}+2 a b+b^{2}, \text { where } a=x, \text { and } b=3\)
\((x+3)^{2}=x^{2}+2 \cdot x \cdot 3+3^{2}\)
Simplify:
\(\left(x^{2}+6 x+9\right)\)Rewrite the equation, substituting \(\left(x^{2}+6 x+9\right)\) for \((x+3)^{2}\).
Distribute the 2.
\(y=2 \cdot x^{2}+6 x \cdot 2+9 \cdot 2+1\)Simplify.
\(y=2 x^{2}+12 x+18+1\)Simplify.
\(y=2 x^{2}+12 x+19\)