Question

The path of a dirt bike rider follows during a jump is given by y=-0.35x^2+3x+12 where x is the horizontal distance (in feet) from the edge of the ramp and y is the height(in feet). What is the maximum height of the rider during the jump?​

Answer 1

The highest height that can be reached is
`(129)/(7)` feet.

Explanation:

The height is
`y`.

Therefore, the maximum height is the
`y`-coordinate of the vertex.

When an equation is in standard form,
`y=ax^2+bx+c`, we can use the following formula to compute the
`x`-coordinate of the vertex:

`(-b)/(2a)`.

So let's compare our equation
`y=ax^2+bx+c` to
`y=-0.35x^2+3x+12`, this gives us:

`a=-0.35`

`b=3`

`c=12`.

Let's begin the
`x`-coordinate computation of the vertex:

`(-b)/(2a)`

`(-3)/(2(-0.35))`

`(3)/(2(0.35))`

`(3)/(0.7)`

`(30)/(7)` (Multiply by 10/10)

To find the corresponding
`y`-coordinate of the vertex given the
`x`-coordinate, we just replace
`x` in
`y=-0.35x^2+3x+12` with
`(30)/(7)` giving us:

`y=-0.35((30)/(7))^2+3((30)/(7))+12`

I will put the right hand side into my calculator:

`y=(129)/(7)`

The highest height that can be reached is
`(129)/(7)` feet.