
How do you state the coordinates of the vertex of the parabola whose equation is : $y = 3{\left( {x - 2} \right)^2} + 5$?
Answer
546.3k+ views
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:
Now consider the given parabola equation $y = 3{\left( {x - 2} \right)^2} + 5$.
Simplifying the given parabola equation as given below:
$ \Rightarrow y = 3{x^2} - 12x + 17$
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 $y = 3{x^2} - 12x + 17$, here $a = 3,b = - 12$ and $c = 17$.
Now finding the x-coordinate of the vertex:
$ \Rightarrow x = \dfrac{{ - \left( { - 12} \right)}}{{2\left( 3 \right)}}$
$ \Rightarrow x = 2$
Now to get the y-coordinate of the vertex of the parabola, substitute the value of $x = 2$, in the parabola equation, as shown below:
$ \Rightarrow y = 3{\left( 2 \right)^2} - 12\left( 2 \right) + 17$
$ \Rightarrow y = 3\left( 4 \right) - 24 + 17$
Simplifying the above equation, as given below:
$ \Rightarrow y = 12 - 7$
$\therefore y = 5$
So the vertex of the parabola $y = 3{x^2} - 12x + 17$ is A, which is given by:
$ \Rightarrow A = \left( {2,5} \right)$
This parabola has its axis parallel to y-axis.
The graph of the given parabola is shown below:
Final answer: The vertex of the parabola is $\left( {2,5} \right)$.
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 by simplifying it into its particular standard form.
Complete step-by-step answer:
Now consider the given parabola equation $y = 3{\left( {x - 2} \right)^2} + 5$.
Simplifying the given parabola equation as given below:
$ \Rightarrow y = 3{x^2} - 12x + 17$
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 $y = 3{x^2} - 12x + 17$, here $a = 3,b = - 12$ and $c = 17$.
Now finding the x-coordinate of the vertex:
$ \Rightarrow x = \dfrac{{ - \left( { - 12} \right)}}{{2\left( 3 \right)}}$
$ \Rightarrow x = 2$
Now to get the y-coordinate of the vertex of the parabola, substitute the value of $x = 2$, in the parabola equation, as shown below:
$ \Rightarrow y = 3{\left( 2 \right)^2} - 12\left( 2 \right) + 17$
$ \Rightarrow y = 3\left( 4 \right) - 24 + 17$
Simplifying the above equation, as given below:
$ \Rightarrow y = 12 - 7$
$\therefore y = 5$
So the vertex of the parabola $y = 3{x^2} - 12x + 17$ is A, which is given by:
$ \Rightarrow A = \left( {2,5} \right)$
This parabola has its axis parallel to y-axis.
The graph of the given parabola is shown below:
Final answer: The vertex of the parabola is $\left( {2,5} \right)$.
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 by simplifying it into its particular standard form.
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