Question

draw the graph of polynomial y=-2x^2+5x+3​

Answer 1

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

The given polynomial is

$\red{\bf :\longmapsto\:y = {2x}^{2} + 5x + 3}$

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 or quadratic polynomial ax² + bx + c is given by

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

Here,

$\rm :\longmapsto\:a = 2$

$\rm :\longmapsto\:b = 5$

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

So, vertex of the parabola is given by

$\rm :\longmapsto\:\bf \:Vertex = \bigg( - \dfrac{ 5}{2 \times 2} , \: \dfrac{4(2)(3) - {5}^{2} }{4 \times 2} \bigg)$

$\rm :\longmapsto\:\bf \:Vertex = \bigg( - \dfrac{ 5}{4} , \: - \dfrac{1}{8} \bigg)$

Step :- 2

Point of intersection with x - axis

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

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

$\rm :\longmapsto\: {2x}^{2} + 5x + 3 = 0$

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

$\rm :\longmapsto\:2x(x + 1) + 3(x + 1)$

$\rm :\longmapsto\:(x + 1)(2x + 3) = 0$

$\rm :\longmapsto\:x = - 1 \: \: \: or \: \: \: x = - 1.5$

Hence, the point of intersection with x- axis is (- 1, 0) and ( - 1.5, 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.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - \dfrac{5}{4} & \sf - \dfrac{1}{8} \\ \\ \sf - 1 & \sf 0 \\ \\ \sf - 1.5 & \sf 0 \\ \\ \sf 0 & \sf 3 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.