Question

How do you find the vertex of $y = 2{x^2} - 4x$?

Answer 1

Hint: This problem deals with the conic sections. A conic section is a curve obtained as the intersection of the surface of a cone with a plane. There are three such types of conic sections which are, the parabola, the hyperbola and the ellipse. This problem is regarding one of those conic sections, which is a parabola. The general form of an equation of a parabola is given by ${x^2} = 4ay$.

Complete step-by-step answer:
The graph of the given parabola is shown below:

Now consider the given parabola equation $y = 2{x^2} - 4x$, writing this in its standard form as shown below:
If the parabola is given by $y = a{x^2} + bx + c$, then the x-coordinate of the vertex is given by:
$ \Rightarrow x = \dfrac{{ - b}}{{2a}}$
Here in the given parabola equation $y = 2{x^2} - 4x$, here $a = 2,b = - 4$ and $c = 0$.
Now finding the x-coordinate of the vertex:
$ \Rightarrow x = \dfrac{{ - \left( { - 4} \right)}}{{2\left( 2 \right)}}$
$ \Rightarrow x = 1$
Now to get the y-coordinate of the vertex of the parabola, substitute the value of $x = 1$, in the parabola equation, as shown below:
$ \Rightarrow y = 2{\left( 1 \right)^2} - 4\left( 1 \right)$
Simplifying the above equation, as given below:
$ \Rightarrow y = 2 - 4$
$\therefore y = 2$
So the vertex of the parabola $y = 2{x^2} - 4x$ is A, which is given by:
$ \Rightarrow A = \left( {1, - 2} \right)$
This parabola has its axis parallel to y-axis.

Final answer: The vertex of the parabola is $\left( {1, - 2} \right)$.

Note:
Please note that if the given parabola is ${x^2} = 4ay$, then the vertex of this parabola is the origin $\left( {0,0} \right)$, and there is no intercept for this parabola as there are no terms of x or y. If the equation of the parabola includes any terms of linear x or y, then the vertex of the parabola is not the origin, the vertex has to be found out by simplifying it into its particular standard form.