Question

Find the equation of the quadratic function f whose minimum value is 2, its graph has an axis of symmetry given by the equation x=−3 and f(2)=1
A. $f(x)={{x}^{2}}+6x-17$
B. $f(x)={{x}^{2}}-6x+17$
C. $f(x)={{x}^{2}}+3x-16$
D. $f(x)={{x}^{2}}+6x-16$
E. Does not exist

Answer 1

Hint: To find the equation of the quadratic function f whose minimum value is 2, its graph has an axis of symmetry given by the equation x=−3 and $f\left( 2 \right)=1$ , we will consider the standard form $f\left( x \right)=a{{x}^{2}}+bx+c$ , where c is a constant. Hence, $f\left( 2 \right)=1=c$ . We know that a vertex of parabola is denoted as $\left( \dfrac{-b}{2a},\dfrac{4ac-{{b}^{2}}}{4a} \right)=\left( h,k \right)$ , $x=h=-3$ and $k={{y}_{\min }}=2$ .Since the axis of symmetry given by the equation x=−3 . Hence, we will get $b=6a$ . Substitute $x=-3$ in $f\left( x \right)=2=a{{x}^{2}}+bx+c$ . We will get $a=-\dfrac{1}{9}$ . Substitute these values in $y=f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ and find $f\left( 2 \right)$ . Check whether $f\left( 2 \right)=1$ or not.

Complete step by step answer:
We have to find the equation of the quadratic function f whose minimum value is 2, its graph has an axis of symmetry given by the equation x=−3 and $f\left( 2 \right)=1$ .
We know that a function is given in the form $f\left( x \right)=a{{x}^{2}}+bx+c$, where c is a constant.
It is given that $f\left( 2 \right)=1$ , that is $f\left( 2 \right)=1=c$
We know that a vertex of parabola is denoted as $\left( \dfrac{-b}{2a},\dfrac{4ac-{{b}^{2}}}{4a} \right)=\left( h,k \right)$
Since the axis of symmetry given by the equation x=−3 , it means that $x=h=-3$ .
Hence, we will get
$\begin{align}
  & \dfrac{-b}{2a}=-3 \\
 & \Rightarrow b=6a...(i) \\
\end{align}$
We have, $f\left( x \right)=a{{x}^{2}}+bx+c$ . Let us substitute the values. We will get
$f\left( x \right)=a{{\left( -3 \right)}^{2}}+b\times -3+1$
By solving this, we will get
$f\left( x \right)=9a-3b+1$
It is given that \[{{y}_{\min }}=2\] . Hence, we can write
$2=9a-3b+1$
Let us now solve this. We will get
$\begin{align}
  & 9a-3b=2-1 \\
 & \Rightarrow 3\left( 3a-b \right)=1 \\
 & \Rightarrow 3a-b=\dfrac{1}{3} \\
\end{align}$
From (i), we can write the above equation as
$\begin{align}
  & 3a-6a=\dfrac{1}{3} \\
 & \Rightarrow -3a=\dfrac{1}{3} \\
 & \Rightarrow a=-\dfrac{1}{9} \\
\end{align}$.
Now, we can write the equation of parabola. We know that the equation of the parabola in vertex form is given as
$y=a{{\left( x-h \right)}^{2}}+k$
Where the vertex is \[\left( h,k \right)=\left( -3,2 \right)\] .
Let us now substitute the values. We will get
$\begin{align}
  & y=\dfrac{-1}{9}{{\left( x-\left( -3 \right) \right)}^{2}}+2 \\
 & \Rightarrow y=\dfrac{-1}{9}{{\left( x+3 \right)}^{2}}+2 \\
\end{align}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Hence,
$y=\dfrac{-1}{9}\left[ {{x}^{2}}+6x+9 \right]+2$
Let us expand this.
$y=\dfrac{-1}{9}{{x}^{2}}-\dfrac{6}{9}x-\dfrac{9}{9}+2$
Let us solve this. We will get
$\begin{align}
  & y=\dfrac{-1}{9}{{x}^{2}}-\dfrac{2}{3}x+1 \\
 & \Rightarrow f(x)=\dfrac{-1}{9}{{x}^{2}}-\dfrac{2}{3}x+1 \\
\end{align}$
Let us find $f\left( 2 \right)$ by substituting $x=2$ in the above equation.
$\Rightarrow f(2)=\dfrac{-1}{9}\times {{2}^{2}}-\left( \dfrac{2}{3}\times 2 \right)+1$
Solving this gives
$f(2)=\dfrac{-4}{9}-\dfrac{4}{3}+1$
Let us take the LCM and solve this.
$\begin{align}
  & f(2)=\dfrac{-4}{9}-\dfrac{4\times 3}{3\times 3}+\dfrac{1\times 9}{1\times 9} \\
 & \Rightarrow f(2)=\dfrac{-4-12+9}{9} \\
 & \Rightarrow f(2)=\dfrac{-7}{9} \\
\end{align}$
It is given that $f\left( 2 \right)=1$ . But we got $f(2)=\dfrac{-7}{9}$ . Hence, no such parabola exists.

So, the correct answer is “Option D”.

Note: You may make an error when writing the formula $y=a{{\left( x-h \right)}^{2}}+k$ as $y=a{{\left( x+h \right)}^{2}}+k$ . This question can also be solved in an alternate way.
Since, the minimum value is given and only 1 axis of symmetry is present, we can assume the function to be a parabola.
It is given that $f\left( 2 \right)=1$ , that is for $x=2,y=1$ .
We can see that the minimum value is 2, that is, $y=2$ . In $f\left( 2 \right)=1$ , $y=1$ which is less than the minimum value $y=2$. Hence, no such parabola exists.