Question

Draw the graph of y= -x2-2x+3 and verify the zeroes using algebraic method

Answer 1

$\large\underline{\sf{Solution-}}$

Given polynomial is

$\rm \: \longmapsto \: y = - {x}^{2} - 2x + 3$

To plot the graph of the quadratic polynomial which is always parabola, the following steps have to be followed :-

Step :- 1 Vertex of parabola

We know, vertex of parabola of quadratic polynomial ax² + bx + c is given by

$\blue{ \boxed{\bf \:Vertex = \bigg( - \dfrac{ b}{2a} , \: \dfrac{4ac - {b}^{2} }{4a} \bigg)}}$

Here,

$\rm :\longmapsto\:a = - 1$

$\rm :\longmapsto\:b = -2$

$\rm :\longmapsto\:c = 3$

So, vertex is

$\rm :\longmapsto\: \:Vertex = \bigg( - \dfrac{ ( - 2)}{2( - 1)} , \: \dfrac{4( - 1)(3) - {( - 2)}^{2} }{4( - 1)} \bigg)$

$\rm :\longmapsto\: \:Vertex = \bigg( - 1 , \: 4\bigg)$

Step :- 2

Point of intersection with x - axis

We know, on x - axis, y = 0.

So,

$\rm :\longmapsto\: - {x}^{2} - 2x + 3 = 0$

$\rm :\longmapsto\:{x}^{2} + 2x - 3 = 0$

$\rm :\longmapsto\:{x}^{2} + 3x - x - 3 = 0$

$\rm :\longmapsto\:x(x + 3) - 1(x + 3) = 0$

$\rm :\longmapsto\:(x + 3)(x - 1) = 0$

$\bf :\longmapsto\:x = - 3 \: \: \: \: or \: \: \: \: x = 1$

Hence,

The point of intersection with x- axis is (1, 0) and ( - 3, 0).

Now,

Point of intersection with y - axis.

We know, on y - axis, x = 0

So, on Substituting the value in given curve, we get

$\rm \: \longmapsto \: y = - {0}^{2} - 2(0)+ 3$

$\rm :\longmapsto\:y = 3$

Hence, the point of intersection with y- axis is (0, 3).

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 3 \\ \\ \sf 1 & \sf 0 \\ \\ \sf - 3 & \sf 0\\ \\ \sf - 1 & \sf 4 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now,

Verification Of zeroes,

From graph, we concluded that zeroes of the given polynomial are - 3 and 1.

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\: - 3 + 1 = - \dfrac{( - 2)}{( - 1)}$

$\rm :\longmapsto\: - 2 = - 2$

Hence, Verified

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\: - 3 \times 1 = \dfrac{3}{ - 1}$

$\rm :\longmapsto\: - 3 = - 3$

Hence, Verified