Algebra - Parabolas

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Section 4.2 : Parabolas

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2. Sketch the graph of the following parabola. The graph should contain the vertex, the \(y\)‑intercept, \(x\)-intercepts (if any) and at least one point on either side of the vertex.

\[f\left( x \right) = 5{\left( {x - 1} \right)^2} - 20\]

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Start Solution

Let’s find the vertex first. Because the parabola is in the form \(f\left( x \right) = a{\left( {x - h} \right)^2} + k\) we know that the vertex is just the point \(\left( {h,k} \right)\). Therefore, we can compare our equation to this form and see that the vertex is : \(\left( {1, - 20} \right)\).

Be careful with minus signs here! For the \(h\) the term the general form is \({\left( {x - h} \right)^2}\) and so to match our equation we must have \(h = 1\) . Likewise, for the \(k\) in the general form it is “+\(k\)” and so to match our equation we need \(k = - 20\) .

Also note that \(a = 5 > 0\) for this parabola and so the parabola will open upwards.

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The \(y\)-intercept is just the point \(\left( {0,f\left( 0 \right)} \right)\). A quick function evaluation gives us that \(f\left( 0 \right) = - 15\) and so for our equation the \(y\)-intercept is \(\left( {0, - 15} \right)\).

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For the \(x\)-intercepts we just need to solve the equation \(f\left( x \right) = 0\). So, let’s solve that for our equation.

\[\begin{align*}5{\left( {x - 1} \right)^2} - 20 & = 0\\ 5{\left( {x - 1} \right)^2} & = 20\\ {\left( {x - 1} \right)^2} & = 4\\ x - 1 & = \pm 2\\ x & = 1 \pm 2 = - 1,3\end{align*}\]

The two \(x\)-intercepts for this parabola are then : \(\left( { - 1,0} \right)\) and \(\left( {3,0} \right)\).

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Because we had two \(x\)-intercepts for this parabola we already have at least one point on either side of the vertex and so we don’t really need to find any more points for our graph.

However, just for the practice let’s find the point corresponding to the \(y\)-intercept on the other side the vertex.

The \(y\)-intercept is a distance of 1 to the left of the vertex and so there will be a corresponding point at the same \(y\) value to the right and it will be a distance of 1 to the right of the vertex. Therefore, the point to the right of the vertex corresponding to the \(y\)‑intercept is \(\left( {2, - 15} \right)\).

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Here is a sketch of the parabola including all the points we found above.