Question

Graph of quadratic polynomial for cases a>0 f(x) = x²-2x-8 and a<0 f(x) = -x²-2x+3

Answer 1

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

Consider,

$\rm :\longmapsto\:f(x) = {x}^{2} - 2x - 8$

Let assume that,

$\rm :\longmapsto\:y = {x}^{2} - 2x - 8$

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,

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

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

$\rm :\longmapsto\:c = - 8$

So,

$\rm :\longmapsto\:\:Vertex = \bigg( - \dfrac{ b}{2a} , \: \dfrac{4ac - {b}^{2} }{4a} \bigg)$

$\rm :\longmapsto\:\:Vertex = \bigg( - \dfrac{ ( - 2)}{2(1)} , \: \dfrac{4(1)( - 8) - {( - 2)}^{2} }{4(1)} \bigg)$

$\rm :\longmapsto\:\:Vertex = \bigg( 1 , \: \dfrac{ - 32 - 4 }{4} \bigg)$

$\rm :\longmapsto\:\:Vertex = \bigg( 1 , \: \dfrac{ - 36 }{4} \bigg)$

$\rm :\longmapsto\:\:Vertex = \bigg( 1 , \: - 9 \bigg)$

Step :- 2

Point of intersection with x - axis

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

So, on substituting the value, we get

$\rm :\longmapsto\:{x}^{2} - 2x - 8 = 0$

$\rm :\longmapsto\:{x}^{2} - 4x + 2x - 8 = 0$

$\rm :\longmapsto\:x(x - 4) + 2(x - 4) = 0$

$\rm :\longmapsto\:(x - 4)(x +2) = 0$

$\bf\implies \:x = 4 \: \: or \: \: x = - 2$

Hence, the point of intersection with x- axis is (4, 0) and ( - 2, 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

$\rm :\longmapsto\:y = {0}^{2} - 2(0) - 8$

$\rm :\longmapsto\:y = - 8$

Hence, the point of intersection with y- axis is (0, - 8).

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 8 \\ \\ \sf 4 & \sf 0 \\ \\ \sf - 2 & \sf 0\\ \\ \sf 1 & \sf - 9 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, Consider

$\rm :\longmapsto\:f(x) = - {x}^{2} - 2x + 3$

Let us assume that,

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

Step :- 1 Vertex of parabola

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

Here,

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

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

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

So,

$\rm :\longmapsto\:\:Vertex = \bigg( - \dfrac{ b}{2a} , \: \dfrac{4ac - {b}^{2} }{4a} \bigg)$

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

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

$\rm :\longmapsto\:\:Vertex = \bigg( - 1 , \: \dfrac{ - 16}{ - 4} \bigg)$

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

Step :- 2

Point of intersection with x - axis

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

So, on substituting the value, we get

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

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

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

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

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

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

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

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

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.

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

➢ Now draw a graph using the points.

➢ See the attachment graph.

Answer 2

Answer:

the above answer is correct.