Algebra - Parabolas

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Section 4.2 : Parabolas

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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.

$f\left( x \right) = 2{x^2} - 12x + 26$

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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,

$\left( { - \frac{{ - 12}}{{2\left( 2 \right)}},f\left( { - \frac{{ - 12}}{{2\left( 2 \right)}}} \right)} \right) = \left( {3,f\left( 3 \right)} \right) = \left( {3,8} \right)$

Also note that $a = 2 > 0$ for this parabola and so the parabola will open upwards.

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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 $f\left( 0 \right) = 26$ and so for our equation the $y$ -intercept is $\left( {0,26} \right)$ .

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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.

$2{x^2} - 12x + 26 = 0\,\,\,\,\, \to \,\,\,\,\,\,x = \frac{{12 \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 2 \right)\left( {26} \right)} }}{{2\left( 2 \right)}} = \frac{{12 \pm \sqrt { - 64} }}{4} = 3 \pm 2i$

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.

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In 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 $\left( {6,26} \right)$ .

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