Section 4.2 : Parabolas
5. 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.
\[f\left( x \right) = 2{x^2} - 12x + 26\]Show All Steps Hide All Steps
Start SolutionLet’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,
\[\left( { - \frac{{ - 12}}{{2\left( 2 \right)}},f\left( { - \frac{{ - 12}}{{2\left( 2 \right)}}} \right)} \right) = \left( {3,f\left( 3 \right)} \right) = \left( {3,8} \right)\]Also note that \(a = 2 > 0\) for this parabola and so the parabola will open upwards.
Show Step 2The \(y\)-intercept is just the point \(\left( {0,f\left( 0 \right)} \right)\). A quick function evaluation gives us that \(f\left( 0 \right) = 26\) and so for our equation the \(y\)-intercept is \(\left( {0,26} \right)\).
Show Step 3For the \(x\)-intercepts we just need to solve the equation \(f\left( x \right) = 0\). So, let’s solve that for our equation.
\[2{x^2} - 12x + 26 = 0\,\,\,\,\, \to \,\,\,\,\,\,x = \frac{{12 \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 2 \right)\left( {26} \right)} }}{{2\left( 2 \right)}} = \frac{{12 \pm \sqrt { - 64} }}{4} = 3 \pm 2i\]So, in this case the solutions to this equation are complex numbers and so we know that this parabola will have no \(x\)-intercepts.
Note that we did not really need to solve the equation above to see that there would be no \(x\)-intercepts for this problem. An alternate method would be to do the following analysis.
From the first step we found that the vertex was \(\left( {3,8} \right)\), which is above the \(x\)-axis, and we also noted that the parabola opened upwards. So, the parabola starts above the \(x\)-axis and opens upwards and we know that once a parabola starts opening in a given direction it won’t turn around and start going in the opposite direction. Therefore, because there is no way for the parabola to go down to the \(x\)-axis, there is no way for there to be \(x\)‑intercepts for this problem.
Show Step 4In this case all we have are the vertex and the \(y\)-intercept (which is on the left side of the vertex). So, we’ll need a point that is on the right 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 3 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 3 to the right of the vertex. Therefore, the point to the right of the vertex corresponding to the y‑intercept is \(\left( {6,26} \right)\).
Show Step 5Here is a sketch of the parabola including all the points we found above.
