How do you write a quadratic function in vertex form whose graph has the vertex \[\left( { - 3,5} \right)\] and passes through the point \[\left( {0, - 14} \right)\]?

Hint: In this question we have to find the quadratic form whose vertex and point through which it passes through, we use the vertex form of the equation which is given by \[y = a{\left( {x - h} \right)^2} + k\], where \[\left( {h,k} \right)\] is the vertex, and now substituting the point through which the equation passes through in the vertex form we will get the value of \[a\], and simplifying the equation we will get the required quadratic equation.

Complete step-by-step solution:
Given that the graph has the vertex \[\left( { - 3,5} \right)\] and it passes through the point \[\left( {0, - 14} \right)\],
We know that the vertex form of the graph is given by \[y = a{\left( {x - h} \right)^2} + k\], where \[\left( {h,k} \right)\] is the vertex,
Now given the vertex of the given function is \[\left( { - 3,5} \right)\],
Comparing the vertex form we can say that,\[h = - 3\] and \[k = 5\],
Now substituting the values in the vertex form we get,
\[ \Rightarrow y = a{\left( {x - \left( { - 3} \right)} \right)^2} + 5\],
Now simplifying we get,
\[ \Rightarrow y = a{\left( {x + 3} \right)^2} + 5\],
So, the equation passes through the point \[\left( {0, - 14} \right)\], now here \[x = 0\] and \[y = - 14\],
Substituting the values in the equation we get,
\[ \Rightarrow - 14 = a{\left( {0 + 3} \right)^2} + 5\],
Now simplifying we get,
\[ \Rightarrow - 14 = a{\left( 3 \right)^2} + 5\],
Now taking the square we get,
\[ \Rightarrow - 14 = a\left( 9 \right) + 5\],
So, now subtracting 5 to both sides we get,
\[ \Rightarrow - 14 - 5 = 9a + 5 - 5\],
Now eliminating the like terms,
\[ \Rightarrow 9a = - 19\],
Now dividing both sides with 9 we get,
\[ \Rightarrow \dfrac{{9a}}{9} = \dfrac{{ - 19}}{9}\],
Now simplifying we get,
\[ \Rightarrow a = \dfrac{{ - 19}}{9}\],
So, by substituting the values in the vertex form, i.e., \[a = \dfrac{{ - 19}}{9}\],\[h = - 3\] and \[k = 5\] then the quadratic function will be,
\[ \Rightarrow \]\[y = \dfrac{{ - 19}}{9}{\left( {x + 3} \right)^2} + 5\],
Now taking L.C.M on the right hand side we get,
\[ \Rightarrow y = \dfrac{{ - 19{{\left( {x + 3} \right)}^2} + 45}}{9}\],
Now taking 9 to the left hand side we get,
\[ \Rightarrow 9y = - 19{\left( {x + 3} \right)^2} + 45\],
Now taking the square we get,
\[ \Rightarrow 9y = - 19\left( {{x^2} + 6x + 9} \right) + 45\],
Now multiplying the terms in the right hand side we get,
\[ \Rightarrow 9y = - 19{x^2} - 114x - 171 + 45\],
Now simplifying we get,
\[ \Rightarrow 9y = - 19{x^2} - 114x - 126\].
So, the quadratic function is \[9y = - 19{x^2} - 114x - 126\].

\[\therefore \]The quadratic function in vertex form whose graph has the vertex \[\left( { - 3,5} \right)\] and passes through the point \[\left( {0, - 14} \right)\] will be equal to \[9y = - 19{x^2} - 114x - 126\].

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.