Question

Draw the Graph for the polynomial
P (x) = x² -3x +2 and find the zeros


from graph?​

Answer 1

Answer:

Method 1:

Prepare a table of value to y=x

2

−x−2.

x −2 −1 0 1 2

y 4 0 −2 −2 0

Draw the parabola. Write the coordinates of the intersecting point of the parabola with x-axis.

Note the co-ordinates of the intersecting points of the parabola with x-axis.

The coordinates where the parabola intersects the x-axis are the roots of the equation.

That coordinates are B(−1,0) and E(2,0).

∴ The roots are x=−1 and x=2

Method 2:

x

2

−x−2=0⇒x

2

=x+2

Consider y=x

2

and y=x+2

For y=x

2

x −2 −1 0 1 2

y 4 1 0 1 4

For y=x+2

x 0 1 −1 2 −2

y 2 3 1 4 0

Draw the parabola and linear graph

Write the coordinates of the points where the straight line intersects with the parabola.

They are B(−1,1) and E(2,4).

∴ The roots are x=−1 and x=2.

solution

Answer 2

Explanation:

Method 1:

Prepare a table of value to y=x

2

−x−2.

x −2 −1 0 1 2

y 4 0 −2 −2 0

Draw the parabola. Write the coordinates of the intersecting point of the parabola with x-axis.

Note the co-ordinates of the intersecting points of the parabola with x-axis.

The coordinates where the parabola intersects the x-axis are the roots of the equation.

That coordinates are B(−1,0) and E(2,0).

∴ The roots are x=−1 and x=2

Method 2:

x

2

−x−2=0⇒x

2

=x+2

Consider y=x

2

and y=x+2

For y=x

2

x −2 −1 0 1 2

y 4 1 0 1 4

For y=x+2

x 0 1 −1 2 −2

y 2 3 1 4 0

Draw the parabola and linear graph

Write the coordinates of the points where the straight line intersects with the parabola.

They are B(−1,1) and E(2,4).

∴ The roots are x=−1 and x=2.

solution

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