Example 1
Completion requirements
| Example 1 |
Determine the equation of the quadratic function graphed below. Write the equation in vertex form.
Step 1: Write down what you know.
vertex is \((-1, -5)\)
\(y\)-intercept is \((0, -2) \)
Step 2: Substitute the values of \(p\) and \(q\) into the vertex form of a quadratic function.
Given the vertex is \((-1, -5)\), \(p = -1\) and \(q = -5\).
\(f(x) = a(x - p)^2 + q \)
\(f(x) = a(x - (-1))^2 + (-5) \)
\(f(x) = a(x + 1)^2 - 5\)
Step 3: Solve for \(a\).
To determine \(a\), use the second point, \((0, -2)\). Substitute \(x = 0\) and \(f(0) = -2\) into the equation, and solve for \(a\).
Please stop at this point and ask yourself if this makes sense…. Does \(a = 3\) work? Yes, it does; the parabola opens upward, and this indicates a positive value for \(a\).
Write the equation of the function.
\(\begin{array}{l}
f\left( x \right) = a\left( {x + 1} \right)^2 - 5 \\
f\left( x \right) = 3\left( {x + 1} \right)^2 - 5 \end{array}\)
vertex is \((-1, -5)\)
\(y\)-intercept is \((0, -2) \)
Step 2: Substitute the values of \(p\) and \(q\) into the vertex form of a quadratic function.
Given the vertex is \((-1, -5)\), \(p = -1\) and \(q = -5\).
\(f(x) = a(x - p)^2 + q \)
\(f(x) = a(x - (-1))^2 + (-5) \)
\(f(x) = a(x + 1)^2 - 5\)
Step 3: Solve for \(a\).
To determine \(a\), use the second point, \((0, -2)\). Substitute \(x = 0\) and \(f(0) = -2\) into the equation, and solve for \(a\).
\(\begin{align}
f\left( x \right) &= a\left( {x + 1} \right)^2 - 5 \\
-2 &= a\left( {0 + 1} \right)^2 - 5 \\
3 &= a\left( 1 \right)^2 \\
3 &= a \end{align}\)
f\left( x \right) &= a\left( {x + 1} \right)^2 - 5 \\
-2 &= a\left( {0 + 1} \right)^2 - 5 \\
3 &= a\left( 1 \right)^2 \\
3 &= a \end{align}\)
Please stop at this point and ask yourself if this makes sense…. Does \(a = 3\) work? Yes, it does; the parabola opens upward, and this indicates a positive value for \(a\).
Write the equation of the function.
\(\begin{array}{l}
f\left( x \right) = a\left( {x + 1} \right)^2 - 5 \\
f\left( x \right) = 3\left( {x + 1} \right)^2 - 5 \end{array}\)