Question

draw the graph of p(x)=x2-x-12 ​

Answer 1

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

Given quadratic polynomial is

$\rm :\longmapsto\:p(x) = {x}^{2} - x - 12$

Let assume that

$\rm :\longmapsto\:y = {x}^{2} - x - 12$

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,

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

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

$\rm :\longmapsto\:c = - 12$

So, on substituting the values, we get

$\rm :\longmapsto\: \:Vertex = \bigg( - \dfrac{( - 1)}{2(1)} , \: \dfrac{4(1)( - 12) - {( - 1)}^{2} }{4(1)} \bigg)$

$\rm :\longmapsto\: \:Vertex = \bigg( \dfrac{1}{2} , \: \dfrac{ - 49 }{4} \bigg)$

$\rm :\longmapsto\: \:Vertex = \bigg(\: \dfrac{1}{2} , \: - \: \dfrac{49 }{4} \bigg)$

Step :- 2

Point of intersection with x - axis

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

So,

$\rm :\longmapsto\: {x}^{2} - 4x + 3x - 12 = 0$

$\rm :\longmapsto\:x(x - 4) + 3(x - 4) = 0$

$\rm :\longmapsto\:(x - 4)(x + 3) = 0$

$\bf\implies \:x = 4 \: \: or \: \: x = \: - \: 3$

Hence, the point of intersection with x- axis is (4, 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

$\rm :\longmapsto\:y = {0}^{2} - 0 - 12$

$\rm :\longmapsto\:y \: = \: - \: 12$

Hence, the point of intersection with y- axis is (0, - 12).

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 12 \\ \\ \sf 4 & \sf 0 \\ \\ \sf - 3 & \sf 0\\ \\ \sf \dfrac{1}{2} & \sf - \dfrac{49}{4} \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.