Question

Select the correct statements for the quadratic polynomial $y=-(x-2{)}^2.$

• A. $\text{Vertex}\equiv (2,0)$
• B. $D=0$
• C. The graph of $y=-(x-2{)}^2$ is given as:
  
• D. $\text{Number of distinct real roots}=2$

Answer 1

Answer: (A), (B), (C)

The graph of $y=-(x-2{)}^2$ is given as:


Given: $y=-(x-2{)}^2$
$\Rightarrow y=-x^2+4x-4$
On comparing with standard form of quadratic expression $\text{y}=a{x}^2+bx+c$
we get, $a=-1,b=4,c=-4$ and $D=b^2-4ac=(4{)}^2-4.(-1).(-4)=0$
Here, $a<0\to \text{downward opening parabola}$ also, $D=0\Rightarrow \text{Both the roots are real and equal}$The coordinates of the vertex will be given as:
$(\frac{-b}{2a},\frac{-D}{4a})$
Substituting the values of $a,bc\&D$ in the equation, we get:
$(\frac{-b}{2a},\frac{-D}{4a})\equiv (\frac{-4}{2.(-1)},\frac{-\left(0\right)}{4.(-1)})\equiv (2,0)$
Hence, the coordinates of vertex is given as:
$(2,0).$

Now, we will find the roots of the equation
$y=-x^2+4x-4=0$
$\Rightarrow x^2-4x+4=0$
$\Rightarrow (x-2{)}^2=0$
$\Rightarrow x=2$

Now with the help of given information we can draw the graph as:

Vertex of the parabola is $(2,0)$ and roots are equal i.e. $2$.