Question

Find the equation of the quadratic function $f$ whose maximum value is $-3$, its graph has an axis of symmetry given by the equation $x=2$ and $f\left(0\right)=-9$

• A. $f\left(x\right)=-\frac{3}{2}{(x-2)}^2-3$
• B. $f\left(x\right)=-\frac{3}{2}{(x+2)}^2+3$
• C. $f\left(x\right)=-\frac{3}{2}{(x-2)}^2$
• D. $f\left(x\right)=-\frac{3}{2}{(x-2)}^2+3$

Answer 1

The correct option is A $f\left(x\right)=-\frac{3}{2}{(x-2)}^2-3$

Maximum value and axis of symmetry occurs at vertex of parabola

Vertex of parabola =$(\frac{-b}{2a},\frac{4ac-b^2}{4a})$

Maximum value =y coordinate of vertex of parabola =-3

Axis of symmetry is at x=2$\Rightarrow$-b/2a = 2

$\Rightarrow$(2,3) is the vertex of parabola

See the above diagram for better understanding

And f(0)=-9$\Rightarrow$c=-9

$\Rightarrow -3=4a+2b-9$

$\Rightarrow 4a+2b=6$

$\Rightarrow$but b=-4a

$\Rightarrow 2a-4a=3$

$\Rightarrow a=-3/2$ and b=6

$\therefore equation:-\frac{3}{2}(x-2{)}^2-3$