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?​

Respuesta :

Answer:

The highest height that can be reached is [tex]\frac{129}{7}[/tex] feet.

Step-by-step explanation:

The height is [tex]y[/tex].

Therefore, the maximum height is the [tex]y[/tex]-coordinate of the vertex.

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

[tex]\frac{-b}{2a}[/tex].

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

[tex]a=-0.35[/tex]

[tex]b=3[/tex]

[tex]c=12[/tex].

Let's begin the [tex]x[/tex]-coordinate computation of the vertex:

[tex]\frac{-b}{2a}[/tex]

[tex]\frac{-3}{2(-0.35)}[/tex]

[tex]\frac{3}{2(0.35)}[/tex]

[tex]\frac{3}{0.7}[/tex]

[tex]\frac{30}{7}[/tex] (Multiply by 10/10)

To find the corresponding [tex]y[/tex]-coordinate of the vertex given the [tex]x[/tex]-coordinate, we just replace [tex]x[/tex] in [tex]y=-0.35x^2+3x+12[/tex] with [tex]\frac{30}{7}[/tex] giving us:

[tex]y=-0.35(\frac{30}{7})^2+3(\frac{30}{7})+12[/tex]

I will put the right hand side into my calculator:

[tex]y=\frac{129}{7}[/tex]

The highest height that can be reached is [tex]\frac{129}{7}[/tex] feet.

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