Question

Guillermo is a professional deep water free diver. his altitude (in meters relative to sea level), xxx seconds after diving, is modeled by g(x)=\dfrac{1}{20}x(x-100)g(x)= 20 1 ​ x(x−100)g, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 20, end fraction, x, left parenthesis, x, minus, 100, right parenthesis what is the lowest altitude guillermo will reach?

Answer 1

Answer: The maximum depth that he will reach is -125 meters.

The first thing you should realize is that this is a quadratic equation and the graph will be a parabola.

We can simply the equation to:

y = (1/20)x^2 - 5x

Now, use -b/2a to find the x-value of the vertex which is 50. Then, input 50 back into the equation to get -125 for the maximum depth.

Answer 2

Answer:

-125 meters.

Step by step explanation:  

We have been given that Guillermo is a professional deep water free diver. his altitude (in meters relative to sea level), x seconds after diving, is modeled by $g(x)=\frac{1}{20}x(x-100)$.

We have been given a quadratic function and the minimum value will occur at vertex.

First of all we will write it in standard quadratic form by using distribution property.

$\frac{x^{2}}{20}-\frac{100x}{20}$

$\frac{x^{2}}{20}-5x$  

Now we will find vertex of our parabola using $x=\frac{-b}{2a}$

$x=\frac{-(-5)}{2(\frac{1}{20})}$

$x=\frac{5}{(\frac{2}{20})}$

$x=\frac{5\times20}{2}$

$x=5\times10$

$x=50$

Now let us substitute x=50 in our quadratic function to find lowest altitude of Guilermo.  

$g(50)=\frac{50^{2}}{20}-5\times 50$

$g(50)=\frac{2500}{20}-250$

$g(50)=125-250$

$g(50)=-125$

Therefore, the lowest altitude Guilermo will reach is -125 meters.