Example 2
Completion requirements
| Example 2 |
Expand the general vertex form \(f\left( x \right) = a\left( {x - p} \right)^2 + q\) to the standard form \(f\left( x \right) = ax^2 + bx + c\).
\(\begin{array}{l}
f\left( x \right) = a\left( {x - p} \right)^2 + q \\
f\left( x \right) = a\left( {x^2 - 2px + p^2 } \right) + q \\
f\left( x \right) = ax^2 - 2apx + ap^2 + q \end{array}\)
Notice that
\(\begin{align}
-2ap &= b \\
p &= \frac{{ -b}}{{2a}} \end{align}\)
This will be the \(x\)-coordinate of the vertex, and the \(y\)-coordinate of the vertex will be the value of \(f\left( p \right) = f\left( {\frac{{ -b}}{{2a}}} \right)\).
f\left( x \right) = a\left( {x - p} \right)^2 + q \\
f\left( x \right) = a\left( {x^2 - 2px + p^2 } \right) + q \\
f\left( x \right) = ax^2 - 2apx + ap^2 + q \end{array}\)
Notice that
\(\begin{align}
-2ap &= b \\
p &= \frac{{ -b}}{{2a}} \end{align}\)
This will be the \(x\)-coordinate of the vertex, and the \(y\)-coordinate of the vertex will be the value of \(f\left( p \right) = f\left( {\frac{{ -b}}{{2a}}} \right)\).
| For another example about coordinates of the vertex, see p. 188 of Pre-Calculus 11. Note that this method can be used on questions such as p. 193 #6 and 11. |