The transformation affects the graph of f horizontally. Find how much time it takes for the object to hit the ground on earth. On the moon, how high is the object at that time?
To obtain g, we must stretch the graph of f horizontally by a factor of 1sqrt(6).
On the moon, the object must be dropped from a height of 53 feet.
Practice makes perfect
Let's begin by figuring out how to get g(t)=- 83t^2+10, starting from f(t)=-16t^2+10.
We can see that because both functions have the same constant 10, the transformation is neither a vertical translation nor a vertical stretch.
Both functions also have the factor - t^2, which implies that the transformation is neither a reflection nor a horizontal translation.
Therefore, the transformation must be a horizontal stretch. This means we need to find a constant a such that f( a* t)= g(t).
This means we can say that to obtain g we must horizontally stretch the graph of f by a factor of a= 1sqrt(6). However, let's find out how much time it takes for the object to hit the ground on Earth. To do that, we must solve f(t)=0.
On earth the object takes sqrt(10)4 seconds to hit the ground. Let's find how high would the object be on the moon after t= sqrt(10)4 seconds. To do that, let's find g( sqrt(10)4).
Therefore, on the moon, after sqrt(10)4 seconds the object is at a height of 253 feet above the ground.
Since we want the object to take the same time to hit the ground both on the earth and on the moon, we need to translate the graph of g(t) down by 253 feet.
g(t) = -8/3t^2+10 - 25/3 Translation down
The new function is h(t)=- 83t^2+ 53. This implies that we need to drop the object from a height of 53 feet.
