Question

How do you write $y = - 3{x^2} + 5x - 2$ in vertex form?

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 given equation is $y = - 3{x^2} + 5x - 2$, the graph of the given equation can be obtained.
The equation of the curve looks like a parabola, a parabola has a vertex.
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 $y = - 3{x^2} + 5x - 2$, here $a = - 3,b = 5$ and $c = - 2$.
Now finding the x-coordinate of the vertex:
$ \Rightarrow x = \dfrac{{ - 5}}{{2\left( { - 3} \right)}}$
$ \Rightarrow x = \dfrac{5}{6}$
Now to get the y-coordinate of the vertex of the parabola, substitute the value of $x = \dfrac{5}{6}$, in the parabola equation, as shown below:
$ \Rightarrow y = - 3{x^2} + 5x - 2$
$ \Rightarrow y = - 3{\left( {\dfrac{5}{6}} \right)^2} + 5\left( {\dfrac{5}{6}} \right) - 2$
Simplifying the above equation, as given below:
$ \Rightarrow y = - 3\left( {\dfrac{{25}}{{36}}} \right) + \dfrac{{25}}{6} - 2$
$ \Rightarrow y = - \dfrac{{25}}{{12}} + \dfrac{{25}}{6} - 2$
Simplifying the value of $y$, as shown below:
$ \Rightarrow y = \dfrac{1}{{12}}$
So the vertex of the parabola $y = - 3{x^2} + 5x - 2$ is A, which is given by:
$ \Rightarrow A = \left( {\dfrac{5}{6},\dfrac{1}{{12}}} \right)$
So the vertex form of the parabola is given by:
$ \Rightarrow y = - 3{\left( {x - \dfrac{5}{6}} \right)^2} + \dfrac{1}{{12}}$
$ \Rightarrow \left( {y - \dfrac{1}{{12}}} \right) = - 3{\left( {x - \dfrac{5}{6}} \right)^2}$

Final Answer: The vertex form of the parabola $y = - 3{x^2} + 5x - 2$ is $\left( {y - \dfrac{1}{{12}}} \right) = - 3{\left( {x - \dfrac{5}{6}} \right)^2}$

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 by simplifying it into its particular standard form.