Question

Select the correct statements for the quadratic polynomial $y=(x-1{)}^2$ and draw the graph.

• A. $\text{Roots}=1$
• B. $\text{Vertex}\equiv (1,0)$
• C. Graph of $y=(x-1{)}^2$ is given as:
  
• D. $D=0$

Answer 1

Answer: (A), (B), (D)

$D=0$

Given: $y=(x-1{)}^2$
$\Rightarrow y=x^2-2x+1$
On comparing with standard form of quadratic expression $\text{y}=a{x}^2+bx+c$
we get, $a=1,b=-2,c=1$ and $D=b^2-4ac=(-2{)}^2-\mathrm{4.1.1}=0$
Here, $a>0\to \text{upward opening parabola}$ 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:
$(\frac{-b}{2a},\frac{-D}{4a})\equiv (\frac{-(-2)}{2.1},\frac{-\left(0\right)}{4.1})\equiv (1,0)$
Hence, the coordinates of vertex is given as:
$(1,0).$

Now, we will find the roots of the equation
$y=x^2-2x+1=0$
$\Rightarrow x^2-2x+1=0$
$\Rightarrow (x-1{)}^2=0$
$\Rightarrow x=1$

Now with the help of given information we will draw the graph



Vertex of the parabola is $(1,0)$ and roots are equal i.e. $1$.