#Tangent Lines Tangent lines are quite useful in many situations when
you need to analyze a function. They can also give you a better
understanding of how certain concepts work such as concaves. In order to
create a tangent line you need to know two things: point slope form
\(y - y_1 = m(x - x_1)\) and a fairly
good knowledge of derivativies. Now that you know what you need to get
started, lets talk about what a tangent line actually is. A tangnet line
is a line that intercepts a function at a specific point and has the
same slope as the function does at that point. For example, in the graph
below the function(in blue) has a tangent line(in red) at the
point(0,1).
The tangent line in this graph has the same slope as the function does
at the point where it intercepts the function. Another thing that needs
to be considered is what if the tangent line intercepts multiple points
of the function? Well, this is usually handled by saying that the point
that the tangent line is based off of has the same slope as the tangent
line. Now that we know what a tangent line is lets look at how to
calculate them. In order to find the formula for a tangent line we need
to know what point we are going to base the tangent line off of. This
point needs to be on the function \(f(x)\). We also need to know the slope at
this point. This is usually found by calculating the derivative. To show
how to do this algebraically, let \((x_1,y_1)\) represent the point we want to
base the tangent line off of. Let \(m\)
represent the slope of the function at the point \((x_1,y_1)\). We can find the equation for
the tangent line like this: \[m =
f'(x_1)\] After we know the slope we can find the formula for
the equation representing the tangent line: \[y - y_1 = m(x - x_1)\] As you can see this
is the exact same formula as point slope form, but we had to find the
variables differently than we normally would. Now, lets look at an
example to further our understanding of the concept. In our example
problem we are going to find the tangent line at the point \((0,7)\) in the function \(f(x) = 3x^2 + 1x + 7\) If we plug in the
value 0 we will get the value 7 just like we expect proving that this
function is a valid function. \[f(0) = 3(0)^2
+ 1(0) + 7\] We can also look at the graph to see that this is a
real point:
Now that we know the point exists, we can find the slope at this point.
We can do this by plugging the x value into the derivative. I am not
going to cover how to find derivatives and I am simply going to show you
the derivative. The derivative of this function is \(f'(x) = 6x + 1\). If we plug our x
value(0) into this formula we get the answer 1. This means that the
slope of the function at this point is 1. If we plug this into the point
slope formula we get this formula: \[y - y_1
= m(x - x_1) = y - 7 = 1(x - 0) = y = x + 7\] So we can see from
this that the formula for our tangent line is \(y = x + 7\). We can verify this by graphing
the formula and the graph. The two functions on the same graph look like
this:
As we can see, the functions have the same slope at that point and
they also intercept the same point.