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Question

# For the given quadratic equation y=x2âˆ’3x+2, select the correct option(s).

A
Graph of y=x23x+2 is

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B
vertex(0,2);
yintercept (32,14)
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C
vertex(32,14);
yintercept (0,2)
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D
Graph of y=x23x+2 is

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Solution

## The correct option is C vertex≡(32,−14); y−intercept ≡(0,2)Given quadratic polynomial: y=x2−3x+2 On comparing with the standard form of quadratic expression y=ax2+bx+c we get, a=1,b=−3,c=2 & D=b2−4ac=(−3)2−4⋅1⋅2=1 Since a>0, the graph would be an upward opening parabola. Now, the vertex of the quadratic polynomial is given by: (−b2a,−D4a) Where −b2a=−−32⋅1=32 & −D4a=−14⋅1=−14 ∴vertex ≡(32,−14) Now, y− intercept is given as the value of y at x=0. ⇒y−intercept≡(0,c)≡(0,2) Thus, we can draw the graph of the polynomial as:

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