Question

Draw the graph if y=-2(x-3)^2+8

Answer 1

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

Given curve is

$\rm :\longmapsto\:y = - 2 {(x - 3)}^{2} + 8$

Since, its a quadratic polynomial, so its graph must be a parabola.

Step :- 1

Point of intersection with x- axis

We know, On x - axis, y = 0,

So, given curve reduced to

$\rm :\longmapsto\:0 = - 2 {(x - 3)}^{2} + 8$

$\rm :\longmapsto\: - 8 = - 2 {(x - 3)}^{2}$

$\rm :\longmapsto\:4 ={(x - 3)}^{2}$

$\rm :\longmapsto\:x - 3 = \: \pm \: 2$

$\rm :\implies\:x = 5 \: \: \: or \: \: \: x = 1$

Hence,

Point of intersection with x - axis is (5, 0) and (1, 0).

Step :- 2

Point of intersection with y- axis

We know, On y - axis, x = 0

So, given curve reduced to

$\rm :\longmapsto\:y = - 2 {(0 - 3)}^{2} + 8$

$\rm :\longmapsto\:y = - 2 {(- 3)}^{2} + 8$

$\rm :\longmapsto\:y = - 18 + 8$

$\rm :\implies\:y = - 10$

Hence,

Point of intersection with y - axis is (0, - 10)

Step :- 3

Vertex of parabola

The given curve is

$\rm :\longmapsto\:y = - 2 {(x - 3)}^{2} + 8$

can be rewritten as

$\rm :\longmapsto\:y - 8 = - 2 {(x - 3)}^{2}$

$\rm :\longmapsto\:{(x - 3)}^{2} = - \: \dfrac{1}{2} (y - 8)$

So, Vertex of parabola is (3, 8).

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 10 \\ \\ \sf 1 & \sf 0 \\ \\ \sf 5 & \sf 0 \\ \\ \sf 3 & \sf 8\end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , - 10), (1 , 0) , (5, 0) & (3 , 8)

➢ See the attachment graph.