Paul's Online Notes
Paul's Online Notes
Home / Algebra / Common Graphs / Parabolas
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 4-2 : Parabolas

6. 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.

\[f\left( x \right) = 4{x^2} - 4x + 1\]

Show All Steps Hide All Steps

Start Solution

Let’s find the vertex first. In this case the equation is in the form \(f\left( x \right) = a{x^2} + bx + c\). And so we know the vertex is the point \(\left( { - \frac{b}{{2a}},f\left( { - \frac{b}{{2a}}} \right)} \right)\). The vertex is then,

\[\left( { - \frac{{ - 4}}{{2\left( 4 \right)}},f\left( { - \frac{{ - 4}}{{2\left( 4 \right)}}} \right)} \right) = \left( {\frac{1}{2},f\left( {\frac{1}{2}} \right)} \right) = \left( {\frac{1}{2},0} \right)\]

Also note that \(a = 4 > 0\) for this parabola and so the parabola will open upwards.

Show Step 2

The \(y\)-intercept is just the point \(\left( {0,f\left( 0 \right)} \right)\). A quick function evaluation gives us that \(f\left( 0 \right) = 1\) and so for our equation the \(y\)-intercept is \(\left( {0,1} \right)\).

Show Step 3

For the \(x\)-intercepts we would normally solve the equation \(f\left( x \right) = 0\). However, in this case we don’t need to do that. From the first step we see that the vertex has a y‑coordinate of zero and hence is also an \(x\)-intercept. Also, because it is the vertex this can be the only \(x\)-intercept for this function.

Note that if we’d solved the equation we would have also arrived at this single \(x\)‑intercept.

Show Step 4

In this case all we have are the vertex (which also happens to be the single \(x\)‑intercept) and the \(y\)-intercept (which is on the right side of the vertex). So, we’ll need a point that is on the left side of the vertex and we can find the point on the left side of the vertex that corresponds to the \(y\)-intercept for this point.

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 \(\left( {1,1} \right)\).

Show Step 5

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