Select the correct statements for the quadratic polynomial y=(x−1)2 and draw the graph.
A
Graph of y=(x−1)2 is given as:
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B
Vertex≡(1,0)
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C
Roots=1
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D
D=0
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Solution
The correct option is DD=0 Given: y=(x−1)2 ⇒y=x2−2x+1
On comparing with standard form of quadratic expression y=ax2+bx+c
we get, a=1,b=−2,c=1 and D=b2−4ac=(−2)2−4.1.1=0
Here, a>0→upward opening parabola also, D=0⇒Both the roots are real and equal
The coordinates of the vertex will be given as: (−b2a,−D4a)
Substituting the values of a,bc&D in the equation, we get: (−b2a,−D4a)≡(−(−2)2.1,−(0)4.1)≡(1,0)
Hence, the coordinates of vertex is given as: (1,0).
Now, we will find the roots of the equation y=x2−2x+1=0 ⇒x2−2x+1=0 ⇒(x−1)2=0 ⇒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.