Question

draw the graph of polynomial f(x)=x^2+2x-3 please it's urgent​

Answer 1

Step-by-step explanation:

Check the graph.

hope this helps :)

Answer 2

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

Given polynomial is

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

Let

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

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 of parabola is

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

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

Step :- 2

Point of intersection with x - axis

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

So, given curve can be

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

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

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

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

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

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

$\rm :\longmapsto\:y = 0 + 0 - 3$

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

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.

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

➢ Now draw a graph using the points.

➢ See the attachment graph.