Question

A flea jumps from the ground to a height of 30 cm and travels 15 cm horizontally from where it started. Suppose the origin is located at the point from which the flea jumped. Determine a quadratic function in vertex form to model the height of the flea compared to the horizontal distance travelled.

Answer 1

Answer:

$y=-\frac{8}{15}x^2+8x$

Step-by-step explanation:

Consider the equation that shows flea's motion is $y=ax^2+bx+c$.

Since the origin is located at the point from where the flea jumped,

So, $y=ax^2+bx+c$  passes through (0,0).

That is, (0,0) must satisfy $y=ax^2+bx+c$,

$0=a(0)^2+b(0)+c$

$\implies c =0$

Now, it travels 15 cm horizontally from where it started.

That is, $y=ax^2+bx+c$  passes through (15,0).

$0=a(15)^2+b(15)+c$

$0=a(225)+b(15)$   (Because c=0)

$225a+15b=0$  ...... (1)

Again, it jumps from the ground to a height of 30 cm,

So, the y-coordinate of vertex of $y=ax^2+bx+c$ 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 = $\frac{15}{2}$ = 7.5,

Thus, vertex = (7.5, 30)

(7.5, 30) must satisfy $y=ax^2+bx+c$

That is, $30=a(7.5)^2+b(7.5)+c$

$30=56.25a+7.5b$

$56.25a+7.5b=30$                ...... (2)

Equation (1) - 2 × equation (2),

225a - 112.5a = -60

112.5a = -60

$\implies a = -\frac{60}{112.5}=-\frac{8}{15}$

From equation (1),

$225(-\frac{8}{15})+15b = 0$

$-120+15b=0$

$15b=120$

$b=8$

Substitute the values of a, b and c in the quadratic equation $y=ax^2+bx+c$,

$y=-\frac{8}{15}x^2+8x$

Which is the required quadratic equation.