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Question: How do you write \(y = - 3{x^2} + 5x - 2\) in vertex form?...

How do you write y=−3x2+5x−2y = - 3{x^2} + 5x - 2 in vertex form?

Explanation

Solution

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 x2=−4ay{x^2} = - 4ay.

Complete step-by-step answer:
The given equation is y=−3x2+5x−2y = - 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=ax2+bx+cy = a{x^2} + bx + c, then the x-coordinate of the vertex is given by:
⇒x=−b2a\Rightarrow x = \dfrac{{ - b}}{{2a}}
Here in the given parabola equation y=−3x2+5x−2y = - 3{x^2} + 5x - 2, here a=−3,b=5a = - 3,b = 5 and c=−2c = - 2.
Now finding the x-coordinate of the vertex:
⇒x=−52(−3)\Rightarrow x = \dfrac{{ - 5}}{{2\left( { - 3} \right)}}
⇒x=56\Rightarrow x = \dfrac{5}{6}
Now to get the y-coordinate of the vertex of the parabola, substitute the value of x=56x = \dfrac{5}{6}, in the parabola equation, as shown below:
⇒y=−3x2+5x−2\Rightarrow y = - 3{x^2} + 5x - 2
⇒y=−3(56)2+5(56)−2\Rightarrow y = - 3{\left( {\dfrac{5}{6}} \right)^2} + 5\left( {\dfrac{5}{6}} \right) - 2
Simplifying the above equation, as given below:
⇒y=−3(2536)+256−2\Rightarrow y = - 3\left( {\dfrac{{25}}{{36}}} \right) + \dfrac{{25}}{6} - 2
⇒y=−2512+256−2\Rightarrow y = - \dfrac{{25}}{{12}} + \dfrac{{25}}{6} - 2
Simplifying the value of yy, as shown below:
⇒y=112\Rightarrow y = \dfrac{1}{{12}}
So the vertex of the parabola y=−3x2+5x−2y = - 3{x^2} + 5x - 2 is A, which is given by:
⇒A=(56,112)\Rightarrow A = \left( {\dfrac{5}{6},\dfrac{1}{{12}}} \right)
So the vertex form of the parabola is given by:
⇒y=−3(x−56)2+112\Rightarrow y = - 3{\left( {x - \dfrac{5}{6}} \right)^2} + \dfrac{1}{{12}}
⇒(y−112)=−3(x−56)2\Rightarrow \left( {y - \dfrac{1}{{12}}} \right) = - 3{\left( {x - \dfrac{5}{6}} \right)^2}

Final Answer: The vertex form of the parabola y=−3x2+5x−2y = - 3{x^2} + 5x - 2 is (y−112)=−3(x−56)2\left( {y - \dfrac{1}{{12}}} \right) = - 3{\left( {x - \dfrac{5}{6}} \right)^2}

Note:
Please note that if the given parabola is x2=−4ay{x^2} = - 4ay, then the vertex of this parabola is the origin (0,0)\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.