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

Answer:

The highest height that can be reached is $\frac{129}{7}$ feet.

Step-by-step 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:

$\frac{-b}{2a}$.

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:

$\frac{-b}{2a}$

$\frac{-3}{2(-0.35)}$

$\frac{3}{2(0.35)}$

$\frac{3}{0.7}$

$\frac{30}{7}$ (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 $\frac{30}{7}$ giving us:

$y=-0.35(\frac{30}{7})^2+3(\frac{30}{7})+12$

I will put the right hand side into my calculator:

$y=\frac{129}{7}$

The highest height that can be reached is $\frac{129}{7}$ feet.