Question

How do you write a quadratic function whose graph has the given characteristics: vertex: \[\left( -3,2 \right)\], point \[\left( 3,12 \right)\]?

Answer 1

Hint: In order to find the solution to the given question that is to find how to write a quadratic function whose graph characteristics are given as vertex: \[\left( -3,2 \right)\], point \[\left( 3,12 \right)\]. Apply the formula of vertex form of a quadratic function which is given by \[y=a{{\left( x-h \right)}^{2}}+k\], where \[\left( h,k \right)\] is the vertex of the parabola. When written in "vertex form": \[\left( h,k \right)\] is the vertex of the parabola, and \[x=h\] is the axis of symmetry. The \[h\] represents a horizontal shift like how far left, or right, the graph has shifted from \[x=0\] and the \[k\] represents a vertical shift like how far up, or down, the graph has shifted from \[y=0\]. A point that the function passes through is \[\left( x,y \right)\]. We have given vertex point \[\left( -3,2 \right)\] that is \[\left( h,k \right)\] and the point \[\left( 3,12 \right)\]that is \[\left( x,y \right)\]. Substitute these values in the \[y=a{{\left( x-h \right)}^{2}}+k\] to find the value of \[a\] and then the quadratic function.

Complete step by step solution:
According to the question, given graph characteristics are as follows:
Vertex: \[\left( -3,2 \right)\] and Point \[\left( 3,12 \right)\].
In order to write a quadratic function, apply the formula of vertex form of the quadratic function which is given by \[y=a{{\left( x-h \right)}^{2}}+k\], where \[\left( h,k \right)\] is the vertex of the parabola and a point that a function passes through is \[\left( x,y \right)\]. We have given vertex point \[\left( -3,2 \right)\] that is \[\left( h,k \right)\] and the point \[\left( 3,12 \right)\] that is \[\left( x,y \right)\]. Substitute these values in the \[y=a{{\left( x-h \right)}^{2}}+k\] to find the value of \[a\], we get:
\[\Rightarrow 12=a{{\left( \left( 3 \right)-\left( -3 \right) \right)}^{2}}+2\]
\[\Rightarrow 12=a{{\left( 6 \right)}^{2}}+2\]
\[\Rightarrow 12=36a+2\]
\[\Rightarrow 10=36a\]
\[\Rightarrow a=\dfrac{10}{36}\]
\[\Rightarrow a=\dfrac{5}{18}\]
Now, substitute the value of \[a\] and the vertex point \[\left( -3,2 \right)\] that is \[\left( h,k \right)\] in the formula \[y=a{{\left( x-h \right)}^{2}}+k\], to find the quadratic function, we get:
\[\Rightarrow y=\dfrac{5}{18}{{\left( x-\left( -3 \right) \right)}^{2}}+2\]
\[\Rightarrow y=\dfrac{5}{18}{{\left( x+3 \right)}^{2}}+2\]
Therefore, the quadratic function whose graph characteristics are given as vertex: \[\left( -3,2 \right)\], point \[\left( 3,12 \right)\] is \[y=\dfrac{5}{18}{{\left( x+3 \right)}^{2}}+2\].

Note: Students can go wrong by interpreting the meaning of \[\left( h,k \right)\] as point that is passing through the function and \[\left( x,y \right)\] as the vertex point, which is completely wrong and leads to the wrong answer. So, key point is to remember that in the formula of vertex form of a quadratic function given by \[y=a{{\left( x-h \right)}^{2}}+k\], where \[\left( h,k \right)\] is the vertex of the parabola and a point that the function passes through is \[\left( x,y \right)\].