Question

Draw a graph of quadratic equation:-. -2x^2+5x+3

Answer 1

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

Given quadratic curve is

$\rm :\longmapsto\:y = - 2 {x}^{2} + 5x + 3 - - (1)$

Note :-

$\rm :\longmapsto\:y = {ax}^{2} + bx + c \: represents \: parabola \: whose \:$

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

So,

For the given quadratic,

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

$\rm :\longmapsto\:vertex \: = \: \bigg( \dfrac{5}{4},\dfrac{49 }{8} \bigg)$

Now,

Substituting 'x = 0' in the given equation (1), we get

$\rm :\longmapsto\:y = - 2 {(0)}^{2} + 5(0) + 3$

$\bf\implies \:y = 3$

Substituting 'x = 3' in the given equation, we get

$\rm :\longmapsto\:y = - 2 {(3)}^{2} + 5(3) + 3$

$\rm :\longmapsto\: y = - 18 + 15 + 3$

$\bf\implies \:y = 0$

Substituting 'x = - 0.5' in the given equation, we get

$\rm :\longmapsto\:y = - 2 {( - 0.5)}^{2} + 5( - 0.5) + 3$

$\rm :\longmapsto\:y = - 0.5 - 2.5 + 3$

$\bf\implies \:y = 0$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 3 & \sf 0 \\ \\ \sf - 0.5 & \sf 0 \\ \\ \sf 0 & \sf 3 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 3), (3 , 0) & (- 0.5 , 0)

➢ See the attachment graph.