Warm Up: Introduction to Standard Form
Completion requirements
Warm Up |
Introduction to Standard Form
\(\begin{array}{l}
f\left( x \right) = - 2\left( {x - 4} \right)^2 + 3 \\
f\left( x \right) = - 2\left( {x - 4} \right)\left( {x - 4} \right) + 3 \\
f\left( x \right) = - 2\left( {x^2 - 4x - 4x + 16} \right) + 3 \\
f\left( x \right) = - 2\left( {x^2 - 8x + 16} \right) + 3 \\
f\left( x \right) = - 2x^2 + 16x - 32 + 3 \\
f\left( x \right) = - 2x^2 + 16x - 29 \\
\end{array}\)
Answers will vary for what you notice. The most obvious, perhaps, is that the leading coefficient, \(a = –2\), is present in both forms.
f\left( x \right) = - 2\left( {x - 4} \right)^2 + 3 \\
f\left( x \right) = - 2\left( {x - 4} \right)\left( {x - 4} \right) + 3 \\
f\left( x \right) = - 2\left( {x^2 - 4x - 4x + 16} \right) + 3 \\
f\left( x \right) = - 2\left( {x^2 - 8x + 16} \right) + 3 \\
f\left( x \right) = - 2x^2 + 16x - 32 + 3 \\
f\left( x \right) = - 2x^2 + 16x - 29 \\
\end{array}\)
Answers will vary for what you notice. The most obvious, perhaps, is that the leading coefficient, \(a = –2\), is present in both forms.