When $y$ is given by an explicit formula in terms of $x$, the point $(x_0,y_0)$ is found by evaluating the $y=f(x)$ at $x_0$, and the slope is found by evaluating the derivative $f'(x)$ at $x=x_0$. By taking advantage of the point-slope form of a line

$$y-y_0=m\left (x-x_0\right ),$$ an equation for the tangent line is found. Let’s explore this in the context of an example.

Suppose that we want to approximate $e^{2x}$ near $x=0$. We pick $c=0$ as the point off of which our approximation will be based, and find the equation of the tangent line to $y=e^{2x}$ at $x=0$.

• $y_0=f(0) = \answer [given]{1}$, so the point $(0,1)$ is on the line.
• $f'(x) = \answer [given]{2e^{2x}}$, so the slope is found by noting $m_{tan} = f'(0) = \answer {2}$.

Thus, the equation of the tangent line is

$$\begin{align*} y-y_0 = m_{tan} (x-x_0) \\ y- \answer{1} &=\answer{2}(x-0) \\ y&=2x+1 . \end{align*}$$

If we “zoom in” on the graphs of the function $y=e^{2x}$ and its tangent line at $x=0$, denoted by $l_0(x)=2x+1$, we see the following picture.

From tangent line approximation, we can approximate values of $e^{2x}$ near $x=0$. Visually, we can see this since the graphs are quite close. Computationally, we obtain the approximations by plugging $x$-values into the equation of the tangent line; for instance, we can approximate $e^{.2}$ by noting

$$e^{.2} = f(.1) \approx 2(.1)+1 =1.2.$$

The actual value of $e^{.2}$ to three decimal places is $1.221$, so the simple arithmetic needed to estimate using the tangent line produces a reasonable approximation. As we “zoom out” to a larger viewing window, however, the graphs start to become quite different.

Since evaluating polynomials involves only arithmetic operations, we would like to be able to use them to give better results than the tangent line approximation. Also, polynomials are easy to integrate and differentiate, so it would be nice to use polynomial approximations in applications that involve these operations. This will require that we try to extract the idea from the tangent line approximation in a way that allows us to generalize it.

Revisiting the Tangent Line Approximation

Let’s look for a first degree polynomial of the form

$$p_1(x) = a_0+a_1x,$$ where $a_0$ and $a_1$ are constants that must be determined.

A Quadratic Approximation

While this should not be too surprising, it does allow for us to think of conditions that will allow for higher degree polynomial approximations. Suppose that we want to use a quadratic polynomial $p_2(x)$ of the form

$$p_2(x) = a_0+a_1x+a_2x^2$$

for making estimates. Note that $f(x)=e^{2x}$ is clearly not linear; in fact, it is concave-up on its domain. Note that by drawing tangent lines at different points near $x=0$, the slopes are different, which is roughly what concavity quantifies. Slopes of tangent lines are found from the first derivative, so in order to measure how these slopes are changing, we should look at the derivative of the first derivative. This is really nothing new; we know already that concavity is measured using the second derivative.

We’ll keep the previous two conditions - that $p_2(0) = f(0)$ and $p_2'(0) = f'(0)$ - and also require that $p_2''(0) = f''(0)$. We thus look for look for a polynomial $$p_2(x) = a_0+a_1x+a_2x^2$$ whose coefficients are found by the requirements

By following the previous example, the reader can (and should) verify that we still have $a_0=1$ and $a_1=2$. To find $a_2$, note that

• $f''(x) = 4e^{2x}$, so $f''(0) = \answer {4}$.
• $p_2'(x) = a_1+2a_2x$, so $p_2''(x) = 2a_2$ and thus $p_2''(0) = 2 a_2$.

Thus, the requirement $p_2''(0) =f''(0)$ gives us that $2a_2=4$, or $a_2=2$.

The quadratic approximation is thus

Let’s now explore our approximations. Geometrically, we can interpret the effectiveness of the approximations by looking at their graphs.

We can also explore the approximations quantitatively for a given $x$-value. For instance, if we want to approximate $e^{.2}$, we note that $f(x)=e^{2x}$, so $e^{.2}=f(.1)$. We thus approximate $e^{.2}$ by evaluating the polynomials at $x=.1$.

• $e^{.2} \approx p_1(.1) = 1+2(.1) = 1.2$
• $e^{.2} \approx p_2(.1) = 1+2(.1)+2(.1)^2 = 1.22$

By noting that the actual value to three decimal place is $1.221$, we can see that the quadratic approximation is better!

Higher Order Approximations

We can continue to look for higher degree polynomial approximations. Note that our approximations above require that the function be sufficiently differentiable at the point at which we wish to base the approximation.

We will develop a more computationally efficient method for computing Taylor Polynomials in the next section, but we conclude this section with a question that explores the ideas put forth so far.