In this lesson we will start off by reviewing the idea of tangents with lots of references to graphs. We will learn how we can calculate the **equation** for what the tangent would be at a point along a line.

This is the intro to our new concept of differentiation. Over this unit we will learn the many methods of how to calculate a formula for the slope of a function. Differentiation gives us the tool to find what the slope is at any point along a function. Slopes have indications about many values in the real world, velocity and acceleration is one of them.

Essentially, slope is the rate of change in a graph at a certain point, it is calculated by how much the horizontal axis changes vs how much the vertical axis changes. Normally, for a graph with the horizontal axis as x, and the vertical axis as y, it can be calculated by,

Slope =

change in y

change in x

Below we have Figure 1, the graph of a function, y = f(x)

Using this graph we can put our *x*, *y*, and slope values into a table as shown below. We will call the slope m(x).

Now we will graph the slope, m(x), treating it like a function. This new concept is graphing what the slope is at a given point for a function m(x).

What we have here is a graph that tells us what the slope value is at a given point, *x* for a function. So, if we pick an x-value on our graphed slope here, the m(x) value will correspond to the slope of the **same** x-value graphed in Figure 1 above.

Velocity is related to the slope of a function that represents distance travelled by time.

Below is the graph of a ball thrown into the air. It measures the height of the ball on the y-axis (in feet) and time after the ball has been thrown on the x-axis (in seconds).

Roughly write the table for the slope (velocity) of the line in the graph above at each point along the x-axis. Then sketch the graph that plots the velocity of the ball (y-axis) against time (x-axis). Velocity is equal to the slope at a given time along the graph of distance against time.

First we roughly illustrate our table of time and slope (velocity) at that time,

Remember, some examples of what a slope is, a flat line has a slope of zero, a line that is going up at a 45 degree angle has a slope of 1, a line that is going vertical 2 values for every 1 horizontal value has a slope of 2, a line that goes down 3 vertical values for every 1 horizontal value has a slope of -3. For this table we are roughly guessing our slope values so try and guesstimate what the slope values are at each point.

Now we have our table, graph the slope (velocity) value on the y-axis and the time value on the x-axis.

This graph gives us the velocity of the ball at a given time. Try and visualize what the velocity of the ball will be like, we throw the ball and right after the ball leaves our hand the velocity is high then after a couple of seconds the ball starts to slow down. After 3 seconds the ball reaches its highest point, at this exact point, the ball stops before it begins moving back towards the ground. After that point the ball has a negative velocity as it is moving backwards towards the starting point. Then after six seconds the ball has no velocity as it has hit the ground and stopped moving.

When we are given a function with a formula, we can calculate the slope of the tangent to any point along that line. [Remember the tangent is the straight line that runs at the slope at a given point]. For two points in that function say, (x, f(x)) and (x + h, f(x + h)) we can calculate the slope of the secant line (m_{sec}) between those points, [recalling on our equations for a slope between two points]

m_{sec} = { f(x + h) - f(x)} / { (x + h) - (x) }

m_{sec} =

f(x + h) - f(x)

(x + h) - (x)

As the value of h gets smaller, the two points (x, f(x)) and (x + h, f(x + h)) get closer and closer together, and therefore the slope between them gets closer and closer to the slope of the tangent at the point (x, f(x)). We say that,

The limit of m_{sec} as *h* approaches 0 will be equal to the slope of the point (x, f(x)) - (m_{tan})

So,

m_{tan} = m_{sec} =

f(x + h) - f(x)

(x + h) - (x)

Find the slope of the tangent at the point (3, 9) on the graph of y = f(x) = x^{2} using the formula,

m_{tan} = m_{sec} =

f(x + h) - f(x)

(x + h) - (x)

So, we have the point (3, 9) on the graphed function y = f(x) = x^{2},

For use in our formula,

x = 3 and x + h = 3 + h

f(x) = x^{2} and f(x + h) = (x + h)^{2},

with x = 3 we have,

f(3) = 3^{2} and f(3 + h) = (3 + h)^{2}

So, sub these values into,

m_{tan} = m_{sec} =

f(x + h) - f(x)

(x + h) - (x)

=

(3 + h)^{2} - (3^{2})

(3 + h) - (3)

=

9 + 6h + h^{2} - (9)

h

=

6h + h^{2}

h

Dividing top and bottom by *h* we get:

= (6 + h)

So, we now have a simplified version, the limit of the formula above as h approaches zero, (theoretically calculating it as h = 0) we get,

= 6 + 0 = 6

That is our slope for the point (3, 9) on the function y = f(x) = x^{2}