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Completion requirements
Determine the equation of the quadratic function graphed below.
The vertex of the graph is \((2, 6)\). Another point to use could be either \((-1, 0)\) or \((5, 0)\) (\(x\)-intercepts).
Using the vertex form, substitute \(2\) for \(p\) and \(6\) for \(q\).
\(\begin{array}{l}
f\left( x \right) = a\left( {x - p} \right)^2 + q \\
f\left( x \right) = a\left( {x - 2} \right)^2 + 6 \end{array}\)
Next, use the point \((-1, 0)\), and substitute \(x = -1\) and \(f(-1) = 0\) into the equation to solve for \(a\).
\(\begin{align}
f\left( x \right) &= a\left( {x - 2} \right)^2 + 6 \\
0 &= a\left( { -1 - 2} \right)^2 + 6 \\
- 6 &= a\left( { -3} \right)^2 \\
- 6 &= 9a \\
\frac{{ - 6}}{9} &= a \\
- \frac{2}{3} &= a \end{align}\)
Finally, put it altogether, and write the equation in vertex form.
Using the vertex form, substitute \(2\) for \(p\) and \(6\) for \(q\).
\(\begin{array}{l}
f\left( x \right) = a\left( {x - p} \right)^2 + q \\
f\left( x \right) = a\left( {x - 2} \right)^2 + 6 \end{array}\)
Next, use the point \((-1, 0)\), and substitute \(x = -1\) and \(f(-1) = 0\) into the equation to solve for \(a\).
\(\begin{align}
f\left( x \right) &= a\left( {x - 2} \right)^2 + 6 \\
0 &= a\left( { -1 - 2} \right)^2 + 6 \\
- 6 &= a\left( { -3} \right)^2 \\
- 6 &= 9a \\
\frac{{ - 6}}{9} &= a \\
- \frac{2}{3} &= a \end{align}\)
Finally, put it altogether, and write the equation in vertex form.
\[\begin{array}{l}
f\left( x \right) = a\left( {x - 2} \right)^2 + 6 \\
f\left( x \right) = - \frac{2}{3}\left( {x - 2} \right)^2 + 6 \end{array}\]
f\left( x \right) = a\left( {x - 2} \right)^2 + 6 \\
f\left( x \right) = - \frac{2}{3}\left( {x - 2} \right)^2 + 6 \end{array}\]