Math Core

In Algebra 1 you graphed lines, parabolas and absolute value "V" shapes one at a time. Algebra 2 organizes them into families. Every family has one simplest member, and every other member is that same graph slid, flipped or stretched. Once you can read those moves straight from an equation, you can sketch a graph like $g(x) = -2\sqrt{x - 1} + 3$ in seconds instead of building a table of values.

The parent functions

Definition

Parent function

A parent function is the simplest function in a family. Every other function in the family is a transformation of it: a shift, reflection, stretch or compression of the parent graph.

These are the parent functions you'll use most this year. Learn their shapes and a few key points on each one, because every transformation starts from those points.

Family	Parent	Key points	Domain	Range
Linear	$f(x) = x$	$(0, 0)$ , $(1, 1)$	all real numbers	all real numbers
Quadratic	$f(x) = x^2$	$(0, 0)$ , $(\pm 1, 1)$ , $(\pm 2, 4)$	all real numbers	$y \ge 0$
Cubic	$f(x) = x^3$	$(0, 0)$ , $(1, 1)$ , $(-1, -1)$	all real numbers	all real numbers
Absolute value	$f(x) = \lvert x \rvert$	$(0, 0)$ , $(\pm 1, 1)$	all real numbers	$y \ge 0$
Square root	$f(x) = \sqrt{x}$	$(0, 0)$ , $(1, 1)$ , $(4, 2)$	$x \ge 0$	$y \ge 0$
Reciprocal	$f(x) = \dfrac{1}{x}$	$(1, 1)$ , $(-1, -1)$	$x \ne 0$	$y \ne 0$

y = x^2y = |x|y = x^3y = sqrt(x)Open in grapher →

The parabola and the V both have a lowest point at the origin, called the vertex. The square root graph starts at the origin and only goes right, because you can't take the square root of a negative number. The cubic passes through the origin, falling on the left and rising on the right.

y = xy = 1/xOpen in grapher →

The reciprocal function never touches either axis. Its graph gets closer and closer to the $x$ - and $y$ -axes; those lines are its asymptotes.

Translations: sliding a graph

Adding a number outside the function moves the graph up or down. Adding a number inside, next to $x$ , moves it left or right.

$g(x) = f(x) + k$ shifts the graph of $f$ up $k$ units (down if $k$ is negative).
$g(x) = f(x - h)$ shifts the graph of $f$ right $h$ units (left if $h$ is negative).

Why does $f(x - 3)$ move the graph to the right? Think about the vertex of $y = (x - 3)^2$ . The inside, $x - 3$ , equals $0$ when $x = 3$ , so the output that the parent produced at $x = 0$ now appears at $x = 3$ . Every point arrives $3$ units later.

The dashed parent y = x² and y = (x − 3)² + 1, shifted right 3 and up 1.Open in grapher →

Common mistake

The horizontal shift goes the opposite way from the sign you see. $f(x + 4)$ moves the graph left $4$ , because $x + 4 = x - (-4)$ , so $h = -4$ . Vertical shifts are not reversed: $f(x) + 4$ moves the graph up $4$ .

Worked example: Reading a translation

Describe how the graph of $g(x) = \lvert x + 4 \rvert - 2$ relates to the graph of $f(x) = \lvert x \rvert$ , and give the vertex of $g$ .

Inside the absolute value, $x + 4 = x - (-4)$ , so the graph moves left 4. Outside, $-2$ moves it down 2.

The vertex of the parent is $(0, 0)$ , so the vertex of $g$ is $(0 - 4, 0 - 2) = (-4, -2)$ .

y = |x|y = |x + 4| - 2Open in grapher →

Reflections: flipping a graph

A negative sign also has two possible spots.

$g(x) = -f(x)$ reflects the graph across the $x$ -axis. Every output changes sign, so $(x, y)$ becomes $(x, -y)$ .
$g(x) = f(-x)$ reflects the graph across the $y$ -axis. Every input changes sign, so $(x, y)$ becomes $(-x, y)$ .

y = sqrt(x)y = -sqrt(x)y = sqrt(-x)Open in grapher →

The square root graph makes the difference easy to see. $y = -\sqrt{x}$ still starts at the origin and goes right, but it heads down. $y = \sqrt{-x}$ is only defined when $-x \ge 0$ , that is, when $x \le 0$ , so it goes left.

Stretches and compressions

Multiplying by a number changes the shape.

$g(x) = a\,f(x)$ multiplies every output by $a$ . If $\lvert a \rvert > 1$ , the graph is vertically stretched (pulled away from the $x$ -axis). If $0 < \lvert a \rvert < 1$ , it is vertically compressed (squashed toward the $x$ -axis). If $a$ is negative, the graph is also reflected across the $x$ -axis.
$g(x) = f(bx)$ multiplies every input by $b$ . This horizontally compresses the graph by a factor of $\dfrac{1}{\lvert b \rvert}$ when $\lvert b \rvert > 1$ and stretches it when $0 < \lvert b \rvert < 1$ .

y = |x|y = 3|x|y = 0.5|x|Open in grapher →

Multiplying the absolute value by $3$ makes the V steeper: the point $(1, 1)$ moves to $(1, 3)$ . Multiplying by $\dfrac{1}{2}$ makes it wider: $(2, 2)$ moves to $(2, 1)$ .

For some parents a horizontal change and a vertical change look identical. For example, $\sqrt{4x} = \sqrt{4}\,\sqrt{x} = 2\sqrt{x}$ when $x \ge 0$ , so compressing $\sqrt{x}$ horizontally by a factor of $\dfrac{1}{4}$ gives the same graph as stretching it vertically by $2$ .

The transformation form

For $g(x) = a\,f\big(b(x - h)\big) + k$ :

Part	Effect on the graph of $f$
$h$	shift right $h$ (left if $h < 0$ )
$k$	shift up $k$ (down if $k < 0$ )
$a$	vertical stretch or compression by $\lvert a \rvert$ ; reflect across the $x$ -axis if $a < 0$
$b$	horizontal compression by $\dfrac{1}{\lvert b \rvert}$ ; reflect across the $y$ -axis if $b < 0$

When $b = 1$ , a point $(x, y)$ on $f$ moves to $(x + h,\ a y + k)$ .

Combining transformations

When several transformations act at once, move key points in this order: stretch or reflect first, then shift. The rule $(x, y) \to (x + h,\ ay + k)$ does both in one step.

Worked example: Graphing with key points

Graph $g(x) = -2\sqrt{x - 1} + 3$ . State its domain and range.

Here $f(x) = \sqrt{x}$ , $h = 1$ , $a = -2$ and $k = 3$ . The graph is reflected across the $x$ -axis, stretched vertically by $2$ , then shifted right $1$ and up $3$ .

Move three key points of $\sqrt{x}$ with $(x, y) \to (x + 1,\ -2y + 3)$ :

Parent point	New point
$(0, 0)$	$(1, 3)$
$(1, 1)$	$(2, 1)$
$(4, 2)$	$(5, -1)$

The graph starts at $(1, 3)$ and heads down and to the right.

y = sqrt(x)y = -2sqrt(x - 1) + 3(1, 3)(2, 1)(5, -1)Open in grapher →

Domain: $x - 1 \ge 0$ , so $x \ge 1$ . Range: the starting value is $3$ and the graph only goes down, so $y \le 3$ .

Worked example: Writing the equation

The graph of $f(x) = x^2$ is reflected across the $x$ -axis, compressed vertically by a factor of $\dfrac{1}{2}$ , and shifted left $2$ and up $5$ . Write an equation for the new function $g$ .

Reflect and compress: $a = -\dfrac{1}{2}$ . Left $2$ : $h = -2$ , so the inside is $x + 2$ . Up $5$ : $k = 5$ .

g(x) = -\frac{1}{2}(x + 2)^2 + 5

The vertex is $(-2, 5)$ , and the parabola opens down.

y = x^2y = -0.5(x + 2)^2 + 5(-2, 5)Open in grapher →

Tip

Check a transformed equation by substituting one key point. The vertex $(-2, 5)$ should satisfy $g$ : $-\frac{1}{2}(-2 + 2)^2 + 5 = 5$ . It does.

Worked example: Transforming a function you can't see

The point $(4, -6)$ lies on the graph of a function $f$ . Find the matching point on each graph.

$g(x) = \dfrac{1}{2}f(x + 3) - 1$
$h(x) = f(2x)$

Solutions.

Here $h = -3$ , $a = \dfrac{1}{2}$ and $k = -1$ . The point moves to $\left(4 - 3,\ \dfrac{1}{2}(-6) - 1\right) = (1, -4)$ . Check: $g(1) = \dfrac{1}{2}f(4) - 1 = -3 - 1 = -4$ .
You need $2x = 4$ , so $x = 2$ . Then $h(2) = f(4) = -6$ , and the point is $(2, -6)$ . The $x$ -coordinate was divided by $2$ , a horizontal compression.

Practice

Practice 1

How does the graph of $g(x) = (x - 5)^2$ relate to the graph of $f(x) = x^2$ ?

Practice 2

What is the vertex of $g(x) = \lvert x + 3 \rvert - 4$ ? Enter it as an ordered pair.

Enter a point like (2, -3)

Practice 3

Which function reflects the graph of $f(x) = \sqrt{x}$ across the $y$ -axis and then shifts it up $2$ ?

Practice 4

The graph of $f(x) = x^2$ is shifted left $4$ units and down $1$ unit. Write the new function $g(x)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

What is the domain of $g(x) = \sqrt{x + 5} - 1$ ? Write it as an inequality in $x$ .

Enter an inequality, e.g. x >= 4, -2 < x <= 3, or x < 1 or x > 5

Practice 6

The function $g(x) = -3\lvert x - 2 \rvert + 7$ has a maximum value. What is it?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

The point $(6, 2)$ is on the graph of $f$ . What point must be on the graph of $g(x) = 3f(x - 1) - 4$ ?

Enter a point like (2, -3)

Practice 8

The graph below is a transformation of $f(x) = \lvert x \rvert$ . Its vertex is $(1, -3)$ and it passes through $(3, 1)$ . Write its equation.

A transformed absolute value graph through (1, −3) and (3, 1).Open in grapher →

Enter an expression, e.g. 3x^2 - 2x + 1