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

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 $\left( x,y \right)$.