WPP #12: Unit O Concept 10: Angles of Elevation and Depression

Steven is playing football and is given the chance to kick the ball into a the field goal.


He is at the 20 yd line and is getting ready to kick it up the 15 yd post. When he kicks it, it hits it exactly at the post. At what angle does he have to kick it to be able to hit the post?

When the ball comes back off the pole it falls 15 yds down and lands on the 30 yd line. At what angle is the ball coming at to the floor?

BQ#7: Unit V Concept 1: The Difference Quotients

When you are given a secant line on a graph with ONE point , you can find the other point by using the difference quotients...
How do you find the difference quotients?
Simple!
First look at the function.

When we give this function a point it will look like this 

We don't know exact numbers so we will give the x value x and the y value f(x) because it is a function. As you can see this is the point for the first point of the secant line. We need to be able to find the slope for this line in order to move on with anything else.

We need to know the point on the graph to be able to use this formula. We can get the other ordered pair on the graph by adding in another variable.

H stands for the length that we are missing. We use H to show that x has changed. When we write the next x value it will look like this...

We have to add x+h because that shows that we are adding both segments. The second point isn't just H. It is both x and h combined together and that is why we add them. The same thing for the y value. Now that we have our points, we are able to plug it in the slope formula. 

Next we are able to cancel variables like x's at the bottom

As a result, we have our difference quotients. Taking the derivative of the quotients formula will get you the slope from the secant line. 

f'(x)=2x will be the slope. From here you can use y=mx+b the find the tangent to any point on the function.