Answer:
[tex]y=-\frac{8}{15}x^2+8x[/tex]
Step-by-step explanation:
Consider the equation that shows flea's motion is [tex]y=ax^2+bx+c[/tex].
Since the origin is located at the point from where the flea jumped,
So, [tex]y=ax^2+bx+c[/tex] passes through (0,0).
That is, (0,0) must satisfy [tex]y=ax^2+bx+c[/tex],
[tex]0=a(0)^2+b(0)+c[/tex]
[tex]\implies c =0[/tex]
Now, it travels 15 cm horizontally from where it started.
That is, [tex]y=ax^2+bx+c[/tex] passes through (15,0).
[tex]0=a(15)^2+b(15)+c[/tex]
[tex]0=a(225)+b(15)[/tex] (Because c=0)
[tex]225a+15b=0[/tex] ...... (1)
Again, it jumps from the ground to a height of 30 cm,
So, the y-coordinate of vertex of [tex]y=ax^2+bx+c[/tex] is 30.
Now, the x-coordinate of vertex of the quadratic function must be half of the distance from its starting point to final point.
That is, x-coordinate of the vertex = [tex]\frac{15}{2}[/tex] = 7.5,
Thus, vertex = (7.5, 30)
(7.5, 30) must satisfy [tex]y=ax^2+bx+c[/tex]
That is, [tex]30=a(7.5)^2+b(7.5)+c[/tex]
[tex]30=56.25a+7.5b[/tex]
[tex]56.25a+7.5b=30[/tex] ...... (2)
Equation (1) - 2 × equation (2),
225a - 112.5a = -60
112.5a = -60
[tex]\implies a = -\frac{60}{112.5}=-\frac{8}{15}[/tex]
From equation (1),
[tex]225(-\frac{8}{15})+15b = 0[/tex]
[tex]-120+15b=0[/tex]
[tex]15b=120[/tex]
[tex]b=8[/tex]
Substitute the values of a, b and c in the quadratic equation [tex]y=ax^2+bx+c[/tex],
[tex]y=-\frac{8}{15}x^2+8x[/tex]
Which is the required quadratic equation.
