Question

How do you determine the vertex as an ordered pair when $f\left( x \right) = 3{\left( {x + 2} \right)^2} - 1$?

Answer 1

Hint: The equation in the question represents a parabola and we have to find the vertex of the given parabola equation, and to find the vertex of the parabola, we should know that the equation of the vertex form is \[y = a{\left( {x - h} \right)^2} + k\], where \[\left( {h,k} \right)\] is the vertex of the parabola, now compare the two equations and we will get the required vertex.

Complete step by step solution:
Given the equation is $f\left( x \right) = 3{\left( {x + 2} \right)^2} - 1$, from the equation we are given that the equation represents a parabola and the parabola does not pass through the origin.
Let us take $y = f\left( x \right) = 3{\left( {x + 2} \right)^2} - 1$,
We can say that,
$y = 3{\left( {x + 2} \right)^2} - 1$,
Now to find the vertex of the parabola,
We know that the equation of parabola in the vertex form can be written as \[y = a{\left( {x - h} \right)^2} + k\], where\[\left( {h,k} \right)\]is the vertex of the parabola,
So here the equation is $f\left( x \right) = 3{\left( {x + 2} \right)^2} - 1$,
Now comparing the two equations we get , \[a = 3\],\[h = - 2\] and \[k = - 1\]
So the vertex of the parabola is \[\left( { - 2, - 1} \right)\],
If we plot the parabola, we get the graph as,

Final Answer:
\[\therefore \] The vertex of the parabola $f \left( x \right) = 3{\left( {x + 2} \right)^2} - 1$ will be equal to \[\left( { - 2, - 1} \right)\], and the graph will be

Note: Symmetric points are called the points which are equidistant from the axis of symmetry and lie on the \[x\]-axis and they are calculated as \[x\]-intercepts. To find the vertex of the parabola we can also make use of the standard form of the equation \[y = a{x^2} + bx + c\], where the axis of symmetry or the \[x\]-coordinate is given by \[x = \dfrac{{ - b}}{{2a}}\] and then we will find the value of \[y\]from the equation of the parabola.