Unit 10 · Lesson 3 of 4
Basic graphs and transformations
You'll be able to: recognize the basic graphs and describe, sketch, or write the equation of a shifted, flipped, or stretched version.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.
Find the domain of \(f(x) = \sqrt{x - 2}\).
The radicand must be at least 0: \(x - 2 \ge 0\), so \(x \ge 2\).
Domain: \([2, \infty)\).
What is the slope of the line \(2x + 3y = 6\)?
Solve for \(y\): \(3y = -2x + 6\), so \(y = -\dfrac{2}{3}x + 2\).
The slope is the number in front of \(x\): \(-\dfrac{2}{3}\).
Step 2 of 5The idea
A few graphs come up again and again. Learn their shapes once, and you can draw many others by moving them.
\(y = \sqrt{x}\) starts at \((0, 0)\) and goes right only. \(y = x^3\) is an S-shape through \((0, 0)\). \(y = x\) is a straight line through \((0, 0)\).
A transformation moves or reshapes one of these. Changes outside the function move it up and down. Changes inside, next to \(x\), move it left and right, and they work backwards.
Up or down: \(f(x) + k\) moves the graph up \(k\). \(f(x) - k\) moves it down \(k\).
Left or right: \(f(x - h)\) moves it right \(h\). \(f(x + h)\) moves it left \(h\).
Flips: \(-f(x)\) flips it over the x-axis (upside down). \(f(-x)\) flips it over the y-axis.
Stretch: \(a \cdot f(x)\) with \(a \gt 1\) makes it taller and steeper; with \(0 \lt a \lt 1\), flatter.
Shortcut: in \(y = a(x - h)^2 + k\) or \(y = a|x - h| + k\), the turning point moves from \((0, 0)\) to \((h, k)\).
Why this shows up on the placement test: "which equation matches this graph?" is a favorite multiple-choice question, and it is quick if you know the shifts.
Step 3 of 5Worked examples
Example 1: graph \(g(x) = (x - 3)^2 + 1\)
- The basic graph is \(y = x^2\), with its turning point (the vertex) at \((0, 0)\).
- Inside: \(x - 3\). The inside is zero at \(x = 3\), so move right 3.
- Outside: \(+1\). Move up 1. The vertex lands at \((3, 1)\).
- Nothing multiplies the square, so the shape stays the same. Check a point: \(g(4) = 1^2 + 1 = 2\), one step right and one up from the vertex, just like \(x^2\).
Example 2: graph \(h(x) = -2|x + 2| + 3\)
- The basic graph is \(y = |x|\), a V with its corner at \((0, 0)\).
- Inside: \(x + 2\) is zero at \(x = -2\), so move left 2.
- The 2 in front stretches it: the sides are twice as steep. The minus sign flips it upside down.
- Outside: \(+3\). Move up 3. The corner lands at \((-2, 3)\), and the V opens down.
- Check a point: \(h(-1) = -2|1| + 3 = 1\), so \((-1, 1)\) is on the graph. One step right of the corner, it went down 2: steeper, and flipped.
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went.
What is the vertex of \(y = (x + 4)^2 - 5\)? Type it as a point.
Inside: \(x + 4\) is zero at \(x = -4\), so the graph of \(x^2\) moves left 4.
Outside: \(-5\) moves it down 5. Vertex: \((-4, -5)\).
Move the graph of \(y = |x|\) right 2 and down 3. Write the new function: \(g(x) = \) ?
Right 2 goes inside, with a minus: \(|x - 2|\).
Down 3 goes outside: \(g(x) = |x - 2| - 3\).
Which equation matches this graph?
- \(y = \sqrt{x - 3} - 1\)
- \(y = \sqrt{x + 3} + 1\)
- \(y = \sqrt{x + 3} - 1\)
- \(y = \sqrt{x - 1} + 3\)
The shape is \(\sqrt{x}\), and its starting point moved from \((0, 0)\) to \((-3, -1)\): left 3 and down 1.
Left 3 is \(x + 3\) inside. Down 1 is \(-1\) outside. That is choice (c).
Check with the other point: \(\sqrt{1 + 3} - 1 = 2 - 1 = 1\). Yes, \((1, 1)\) is on it.
The graph of \(y = x^3\) is stretched vertically by a factor of 2, then moved up 4. Write the new function: \(g(x) = \) ?
Stretch by 2: multiply the function by 2, giving \(2x^3\).
Up 4: add 4 outside. \(g(x) = 2x^3 + 4\).
How do you get the graph of \(y = -(x + 1)^2\) from the graph of \(y = x^2\)?
- Shift left 1, then flip over the x-axis.
- Shift right 1, then flip over the x-axis.
- Shift left 1, then flip over the y-axis.
- Shift up 1, then flip over the x-axis.
Inside: \(x + 1\) is zero at \(x = -1\), so shift left 1. That rules out (b) and (d).
The minus sign is outside the square, so the outputs change sign: a flip over the x-axis, and the U opens down. (A y-axis flip would leave it opening up.)
Answer: (a).
The point \((2, 4)\) is on the graph of \(y = x^2\). Where does that point move to on the graph of \(y = (x - 1)^2 + 3\)?
\((x - 1)\) inside means right 1: the x-value goes from 2 to 3.
\(+3\) outside means up 3: the y-value goes from 4 to 7.
New point: \((3, 7)\). Check: \((3 - 1)^2 + 3 = 4 + 3 = 7\).
What is the range of \(y = -|x + 1| + 5\)? Think about the graph.
The V of \(|x|\) moves left 1 and up 5, so its corner is at \((-1, 5)\).
The minus sign flips it, so the V opens down. The corner is the highest point, and the sides go down forever.
Range: every y up to and including 5, \((-\infty, 5]\).
Let \(f(x) = x^2\) and \(g(x) = f(x + 3) - 2\). Write \(g(x)\) as a polynomial: \(g(x) = \) ?
Replace the input of \(f\) with \((x + 3)\): \(g(x) = (x + 3)^2 - 2\). This is \(x^2\) moved left 3 and down 2.
Expand: \((x + 3)^2 = x^2 + 6x + 9\).
Subtract 2: \(g(x) = x^2 + 6x + 7\).
Step 5 of 5Wrap-up
You can now picture \(y = a \cdot f(x - h) + k\) as the basic graph flipped or stretched by \(a\), moved right \(h\), and moved up \(k\).
Read or watch next: OpenStax College Algebra 2e, 3.5 Transformation of Functions (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Algebra 2: Transformations of functions (unit) (short videos). Quick rules: Functions cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why does \(x - 3\) inside the parentheses move the graph right instead of left?" It knows this course and can explain things another way.