Graphs of powers, exponentials, logarithms and absolute value

Know a handful of shapes by heart

Without a calculator, graph questions are answered by recognising a basic shape and a few of its points. Learn these.

Function	Domain	Shape and key points
$y = x^2$	all reals	U-shaped, lowest point $(0, 0)$, symmetric about the $y$-axis; passes $(\pm 1, 1)$, $(\pm 2, 4)$.
$y = x^3$	all reals	Always increasing, flat at the origin, symmetric through the origin; passes $(1, 1)$, $(-1, -1)$, $(2, 8)$.
$y = \sqrt{x}$	$x \geq 0$	Starts at $(0, 0)$, rises ever more slowly; passes $(1, 1)$, $(4, 2)$, $(9, 3)$.
$y = \frac{1}{x}$	$x \neq 0$	Two separate branches, in the first and third quadrants; the axes are asymptotes; passes $(1, 1)$, $(-1, -1)$.
$y = a^x$, $a > 1$	all reals	Always positive, increasing, passes $(0, 1)$ and $(1, a)$; hugs the $x$-axis on the left.
$y = a^x$, $0 < a < 1$	all reals	Always positive, decreasing, passes $(0, 1)$; hugs the $x$-axis on the right.
$y = \log_a x$, $a > 1$	$x > 0$	Increasing, passes $(1, 0)$ and $(a, 1)$; plunges towards $-\infty$ next to the $y$-axis.
$y = \lvert x \rvert$	all reals	V-shaped, corner at $(0, 0)$, slopes $-1$ and $1$.

Two facts link these. The exponential $a^x$ and the logarithm $\log_a x$ are inverse functions, so their graphs are mirror images in the line $y = x$: the point $(0, 1)$ on one becomes $(1, 0)$ on the other, and the horizontal asymptote $y = 0$ becomes the vertical asymptote $x = 0$. And $y = \left(\frac{1}{2}\right)^x$ is $y = 2^{-x}$, the mirror image of $2^x$ in the $y$-axis.

Moving a graph

Start from a known graph $y = f(x)$ and $k > 0$.

New equation	Effect on the graph
$y = f(x) + k$	up by $k$
$y = f(x) - k$	down by $k$
$y = f(x - k)$	right by $k$
$y = f(x + k)$	left by $k$
$y = -f(x)$	reflected in the $x$-axis
$y = f(-x)$	reflected in the $y$-axis
$y = \lvert f(x) \rvert$	parts below the $x$-axis flipped above it

The horizontal shifts feel backwards, and that is exactly why they are tested. In $y = f(x - 3)$, the output that $f$ used to give at $0$ now appears at $x = 3$, so everything has moved right. Check any shift with one point: $y = \sqrt{x - 3}$ starts where the radicand is zero, at $x = 3$.

Asymptotes move with the graph. $y = 2^x - 4$ has the horizontal asymptote $y = -4$; $y = \log_2(x + 1)$ has the vertical asymptote $x = -1$ and the domain $x > -1$.

Absolute value of a function

$y = \lvert f(x) \rvert$ keeps every point where $f$ is positive or zero and reflects the rest upwards. So $y = \lvert x^2 - 4 \rvert$ is the U-shape of $x^2 - 4$ with the stretch between $x = -2$ and $x = 2$, which dipped to $-4$, flipped into an arch that peaks at $(0, 4)$. The whole graph looks like a W: its two lowest points, at $(\pm 2, 0)$, are sharp corners, and its middle peak is rounded. Compare $y = f(\lvert x \rvert)$, which is a different operation: it copies the right half of the graph onto the left.

Counting solutions with graphs

The real solutions of $f(x) = g(x)$ are the $x$-coordinates where the graphs of $f$ and $g$ meet. You often do not need the solutions themselves, only how many there are, and a rough sketch answers that. Two tools make the count reliable:

• Monotonicity. If one side is increasing and the other decreasing, the graphs meet at most once. If you also find one solution, it is the only one.
• Sign changes. If $f - g$ is positive at one point and negative at another, and both are unbroken in between, the graphs cross somewhere between them.

Decreasing on the whole domain

"Decreasing" means that a larger input always gives a smaller output. $y = \frac{1}{x}$ falls on each branch separately, but it is not decreasing on its whole domain: $-1 < 1$ and $\frac{1}{-1} = -1 < 1 = \frac{1}{1}$. The break at $x = 0$ lets the graph jump from $-\infty$ to $+\infty$.