Intersections and Transformations of Graphs

Intersecting the Coordinate Axes

To determine where the graph of a function $y = f(x)$ intersects the coordinate axes, we use specific algebraic substitutions. For the $y$ axis, the $x$ coordinate must be zero. By calculating $f(0)$, we find the $y$ coordinate of the intersection. For the $x$ axis, the $y$ coordinate must be zero. We solve the equation $f(x) = 0$ to find the $x$ coordinates. These solutions are known as the real roots of the function. A general polynomial of degree $n$ can possess a maximum of $n$ real roots, though it may have fewer.

The Notation $y = f(x)$

Before exploring transformations, it is vital to understand the notation. The expression $y = f(x)$ indicates that for any value of $x$, the corresponding $y$ value on the curve is calculated using the function $f$. For example, if $f(x) = x^2 + 3$, then at $x = 2$, $y = 2^2 + 3 = 7$. At $x = 4$, $y = 4^2 + 3 = 19$. This relationship allows us to deduce how changes to the algebra of the function affect the physical graph.

Vertical Stretches: $y = a f(x)$

When we multiply the entire function by a constant $a$, every $y$ value is multiplied by that constant. If we compare $y = x^3$ with $y = 4x^3$, each $y$ value on the second graph is four times as large as on the first. This results in a vertical stretch away from the $x$ axis by a factor of 4.

In general, $y = a f(x)$ stretches the graph vertically by factor $a$. If $0 < a < 1$, the graph becomes less tall (a vertical squash). If $a$ is negative, the graph is reflected in the $x$ axis in addition to the stretch.

Vertical Translations: $y = f(x) + a$

Adding a constant $a$ to the function result moves the entire graph vertically. For $y = x^2 + 3$, every point on the graph $y = x^2$ is shifted up by 3 units. Formally, we describe this as a translation by the vector $\binom{0}{a}$. If $a$ is negative, the graph moves downwards parallel to the $y$ axis.

Note that in trigonometry, we often write $a + \cos x$ instead of $\cos x + a$ to avoid confusion with $\cos(x + a)$.

Horizontal Translations: $y = f(x + a)$

This transformation is frequently misunderstood. While it may seem intuitive that adding $a$ to $x$ should shift the graph to the right, it actually shifts the graph to the left. To find $f(x + a)$ from a given $f(x)$, we replace every instance of $x$ in the expression with $(x + a)$.

Example: Given $f(x) = x^2 + 2x - 5$, then $f(x + 3) = (x + 3)^2 + 2(x + 3) - 5$.

Example: Given $f(x) = \cos(2x)$, then $f(x - \frac{\pi}{2}) = \cos 2(x - \frac{\pi}{2}) = \cos(2x - \pi)$. It is a common error to forget to multiply the entire replacement by the coefficient of $x$.

To understand why $y = f(x + 3)$ shifts the graph of $f(x) = 2^x$ to the left, consider the $y$ values. On $y = f(x)$, the $y$ value above $x = 5$ is $2^5 = 32$. On $y = f(x + 3)$, the $y$ value above $x = 2$ is $f(2+3) = f(5) = 32$. The value that used to occur at $x=5$ now occurs earlier at $x=2$. Thus, the graph has moved 3 units to the left.

In general, $y = f(x + a)$ is a translation of $y = f(x)$ by the vector $\binom{-a}{0}$.

Horizontal Stretches: $y = f(ax)$

Multiplying the input $x$ by a factor $a$ squashes the graph towards the $y$ axis by a factor of $a$. To find $f(ax)$, we replace every $x$ with $ax$. For example, if $f(x) = x^2$, then $f(3x) = (3x)^2 = 9x^2$.

Consider $f(x) = x^3 - 3x^2 + 2$. If we look at $y = f(2x)$, the $y$ value above $x = 1$ is $f(2 \times 1) = f(2)$. In the original graph, the $y$ value for $x = 2$ was $-2$. Now, that same $y$ value occurs at $x = 1$. This causes the graph to squash horizontally.

In general, $y = f(ax)$ is a horizontal stretch by a factor of $\frac{1}{a}$ towards the $y$ axis. If $a < 0$, the graph is also reflected in the $y$ axis.

Combining Transformations

The order in which transformations are applied is critical. Consider $y = \cos(2x + \frac{\pi}{6})$. There are two ways to interpret this starting from $y = \cos x$:

1. Translate by $\binom{-\frac{\pi}{6}}{0}$ to get $\cos(x + \frac{\pi}{6})$, then apply a horizontal squash by factor 2 to get $\cos(2x + \frac{\pi}{6})$. This is correct.

2. Squash by factor 2 to get $\cos 2x$, then translate by $\binom{-\frac{\pi}{12}}{0}$. This results in $\cos 2(x + \frac{\pi}{12}) = \cos(2x + \frac{\pi}{6})$. This is also correct.

If you squash by 2 first and then translate by $\frac{\pi}{6}$, you would get $\cos 2(x + \frac{\pi}{6}) = \cos(2x + \frac{\pi}{3})$, which is incorrect.

The Notation $f(g(x))$

Composite functions involve taking the output of one function as the input for another. To find $f(g(x))$, replace every $x$ in $f(x)$ with the entire expression for $g(x)$. For example, if $g(x) = 2x$ and $f(x) = x^2 + 3x - 2$, then $f(g(x)) = (2x)^2 + 3(2x) - 2 = 4x^2 + 6x - 2$. Generally, $f(g(x))$ is not equal to $g(f(x))$.

Linear and Quadratic Graphs

We can view standard equations through the lens of transformations:

1. Linear: $y = mx + c$ can be seen as $y = x$ stretched vertically by factor $m$ then translated by $\binom{0}{c}$.

2. Quadratic: $y = a(x + b)^2 + c$ is the graph $y = x^2$ stretched vertically by $a$, translated horizontally by $-b$, and translated vertically by $c$. The vertex of this parabola is located at $(-b, c)$.