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 $f\left(2\right)=1$

• A. $f\left(x\right)=x^2+6x-17$
• B. $f\left(x\right)=x^2-6x+17$
• C. $f\left(x\right)=x^2+3x-16$
• D. $f\left(x\right)=x^2+6x-16$
  
• E. Does not exist

Answer 1

The correct option is E Does not exist

Since it is given that the function has a minimum value then it is an upward open parabola. Now its axis of symmetry is $x=-3$. Hence the function achieves a minimum value at $x=-3$. Now the minimum value is given as 2. Thus the vertex of the parabola is $(-3,2)$.
Hence the equation of the parabola is given as
$(x+3{)}^2=y-2$ or $x^2+6x+9=y-2$ or $y=x^2+6x+11$
Hence
$f\left(x\right)=x^2+6x+11$.
Now
$f\left(2\right)=4+12+11=27\ne 1$. Thus there exists no such parabola as given in the question.