From the Graphing Topic earlier, a new form of quadratic functions was introduced, the vertex form. If \((h,k)\) is the vertex of the parabola, then an alternate form equation for the parabola looks like: \(f(x)=a(x-h)^2+k\).
NOTE: It is important to realize that the equation has \(x-h\), so when \(h=-4\) for example, then the equation will show \(x+4\). The value of \(a\) is the same as in the standard form. When that value is not known, you must find it by using another point.
Find an equation in vertex form for the graph below.
From the graph, the vertex is at \((6,4)\). Plugging into the vertex form gives
\(f(x)=a(x-6)^2+4\). The only thing left to do, is find the correct value of \(a\).
This is what the second point from the graph \((5,1)\) is used for, to find \(a\). Plug in \(\color{red}{x=5}\) and \(\color{green}{y=1}\) into the vertex form:$$\begin{array}{l} \color{green}{1}=a(\color{red}{5}-6)^2+4 \\ \implies 1=a(-1)^2+4 \\ \implies 1=a\cdot1+4 \\ \implies1-4=a \implies-3=a\color{green}{\checkmark} \end{array}$$
This gives the vertex form for this graph as \(f(x)=-3(x-6)^2+4\).
One big question arises here, how can a standard equation be transformed into the vertex form equation?
The algebra steps needed are discussed below using \(f(x)=3x^2+24x+50\).
| Description of Steps | Applied to Example |
|---|---|
| Find \(h\) using the formula: \(h=\large{\frac{-b}{2a}}\) | \(a=3\) and \(b=24\implies\begin{array}{l} h=\large{\frac{-24}{2(3)}} \\ \implies h=\large{\frac{-24}{6}} \\ \implies h=-4 \end{array}\) |
| Find \(k\) by plugging in \(h\) for \(x\) in the standard form: | \(k=f(-4)\implies \begin{array}{l} k=3(-4)^2+24(-4)+50 \\ \implies k=3(16)-96+50 \\ \implies k=48-46 \\ \implies k=2 \end{array}\) |
| Locate the value of \(a\) from the standard form and include it in vertex form: | \(a=3\) so the vertex form is \(f(x)=3(x+4)^2+2\) |
How can Desmos help us understand these ideas? By clicking on the graph, it will indentify the vertex with a grey dot at the point. Moving to that point will show the coordinates of the vertex. This is a really nice way to check work after performing the algebra.
The graph shows the vertex located at \((-4,2)\).
That was the same vertex found by doing the algebra in the example above.
So use technology to check for the same result found by the algebra.
This same approach applies to other points identified by Desmos.
The \(x\)-intercepts and the \(y\)-intercept, for example, will also show up on the graph.
From the vertex form equation, a graph can be sketched fairly quickly. Recall that \(a\) tells us if the parabola opens up or down, also about skinny or wide compared to the basic graph. From the Graphing/Standard Form topic, follow the steps to graph. The advantage to vertex form is that step 1 is already done, the vertex is given. Then find the \(y\)-intercept (remember \(x=0\)) and graph that point. Finally, use the axis of symmetry to find the third point.
Sketch a graph of the quadratic function \(f(x)=-3(x+1)^2+6\).