Graphing square root functions
A rule like f(x) = 2√(x − 3) + 1 refuses some inputs. Where does its graph begin, which way does it run, and which heights does it reach?
Where you are 3 in Radical functions
When you reach for it
Reach for this whenever a rule puts its input under a square root sign: \displaystyle f(x) = 2\sqrt{x - 3} + 1, a pendulum’s period as a function of its length, the distance to the horizon as a function of height. Three questions come with every such rule, and this skill answers all three at once: which inputs it accepts, what its graph looks like, and which outputs it can produce.
The graph is never a whole curve across the page. It is a half curve: it begins at a single point, its endpoint, and runs off in one direction only, steep at first and flattening as it goes. Drawing it well means finding that endpoint exactly, knowing which way the curve runs from it, and plotting a few points whose outputs come out as whole numbers.
You already have the pieces. The square root function gave the parent graph of \displaystyle y = \sqrt{x}, through (0, 0), (1, 1), (4, 2) and (9, 3). The implied domain said a square root accepts only inputs that make the expression under it at least 0. And graphing by transformations showed how to move a parent graph. This lesson puts them together for the square root family and adds what is special about it: the endpoint, the four directions, and the rule \displaystyle \sqrt{c^2 x} = c\sqrt{x} that turns a squeeze into a stretch.
The procedure
The running example is the lead rule
The expression under the square root sign, here x - 3, is the radicand.
The domain comes first
No real number squares to a negative, so \displaystyle \sqrt{\;} is defined only for radicands that are at least 0. That is the whole of the domain question:
So the domain is x \ge 3, or [3, \infty) in Interval and set-builder notation. Nothing is drawn to the left of x = 3: there is no graph there, not even a faint one, because there are no outputs there.
The endpoint is where the radicand is zero
At the edge of the domain the radicand is exactly 0, and \displaystyle \sqrt{0} = 0, so everything that depends on the root vanishes and only the constants survive:
The endpoint is (3, 1). In general, for \displaystyle y = a\sqrt{x - h} + k the radicand is 0 at x = h and the output there is a \cdot 0 + k = k, so the endpoint is (h, k). That matches what Graphing by transforming a parent graph said about the parent’s anchor (0, 0): it moves to (h, k).
Choose inputs that make the radicand a perfect square
Any input in the domain gives a point, but most give outputs such as \displaystyle 2\sqrt{2} + 1, which are hard to plot. Choose instead the inputs that make the radicand 0, 1, 4 and 9, because their square roots are 0, 1, 2 and 3:
| x - 3 | x | \displaystyle \sqrt{x - 3} | f(x) |
|---|---|---|---|
| 0 | 3 | 0 | 1 |
| 1 | 4 | 1 | 3 |
| 4 | 7 | 2 | 5 |
| 9 | 12 | 3 | 7 |
The first column is the radicand. Work across each row from it: pick the square, solve for x, then compute the output. This is the same as carrying the parent’s points (0, 0), (1, 1), (4, 2), (9, 3) through the move (p, q) \mapsto (p + 3,\ 2q + 1), done from the other end.
Which way it runs
The domain x \ge 3 lies to the right of the endpoint, so the curve runs right. The outputs are 1 plus twice something that is never negative, so they are never below 1: the curve runs up. A square root graph always runs in one of four directions, set by two signs.
- The sign in front of the root decides up or down. With a > 0 the outputs climb from the endpoint; with a < 0, as in \displaystyle -\sqrt{x}, every output is the negative of a climbing one, so they fall.
- The sign of x inside the root decides right or left. With +x inside, the radicand grows as x grows, so the allowed inputs lie to the right of the endpoint. With -x inside, as in \displaystyle \sqrt{-x}, the radicand grows as x shrinks, so the allowed inputs lie to the left.
Either way the endpoint is an extreme: the lowest point of a curve that runs up, the highest point of one that runs down.
The range comes from the endpoint and the direction
The outputs start at the endpoint’s height and run off in one direction without stopping. For f, the range is y \ge 1, or [1, \infty). Every height from 1 upward really is reached: to reach height M \ge 1, solve \displaystyle 2\sqrt{x - 3} + 1 = M, which gives \displaystyle x = 3 + \left(\frac{M - 1}{2}\right)^2, an input in the domain. In general:
The range does not depend on whether the curve runs left or right. Only the sign in front of the root and the endpoint’s height matter.
The shape between the points
Join the points with a smooth curve that is steep at the endpoint and flattens as it goes, never a straight segment and never a sharp corner. The table shows why it flattens. Each row rises by 2, but the runs between the inputs 3, 4, 7, 12 are 1, 3, 5: equal rises cost longer and longer runs. For the parent, rising from height n to n + 1 takes the inputs from n^2 to (n + 1)^2, a run of 2n + 1, and that run keeps growing.
Yet the curve never levels off. Flattening is not stopping: the curve passes every height eventually, as the range argument showed. Near the endpoint the opposite happens. The parent climbs from 0 to 0.1 while x moves only from 0 to 0.01, so the curve leaves the endpoint almost straight up.
When there is a number in front of x inside
A rule such as \displaystyle \sqrt{4x + 12} does not show its endpoint at a glance. Two moves help. The domain still comes from the radicand, 4x + 12 \ge 0, so x \ge -3. And when the number in front of x is a perfect square, it can be pulled out of the root. For any c > 0 and x \ge 0,
because the square root of a product of non-negative numbers is the product of their square roots. So \displaystyle \sqrt{4x + 12} = \sqrt{4(x + 3)} = 2\sqrt{x + 3}: a horizontal squeeze by 4 draws exactly the same graph as a vertical stretch by 2. Example 4 works one through.
Worked examples
The lead rule: domain, graph and range
Graph \displaystyle f(x) = 2\sqrt{x - 3} + 1 and state its domain and range.
SolutionThe radicand must be at least 0: x - 3 \ge 0, so the domain is x \ge 3, that is [3, \infty). At x = 3 the radicand is 0, so f(3) = 2 \cdot 0 + 1 = 1, and the endpoint is (3, 1).
Next, the inputs that make the radicand 0, 1, 4, 9 are x = 3, 4, 7, 12, and they give the points (3, 1), (4, 3), (7, 5) and (12, 7), as in the table above. For example \displaystyle f(7) = 2\sqrt{4} + 1 = 2 \cdot 2 + 1 = 5.
The number in front of the root is 2 > 0, so the curve runs up, and the inside is x - 3, with +x, so it runs right. Plot the four points and join them with a curve that leaves (3, 1) almost straight up and bends toward the right. Put a filled dot at (3, 1), since that point is on the graph, and an arrow at the far end, since the curve goes on.
The range starts at the endpoint’s height and goes up: y \ge 1, that is [1, \infty).
Compared with the parent \displaystyle y = \sqrt{x}, every height above the endpoint is doubled: the parent rises 1 over the first unit of run, this curve rises 2.
A curve that runs down
Graph \displaystyle g(x) = -\sqrt{x + 4} + 2, state its domain and range, and find where it crosses the axes.
SolutionThe radicand is x + 4, so the domain is x + 4 \ge 0, that is x \ge -4, or [-4, \infty). At x = -4 the output is \displaystyle -\sqrt{0} + 2 = 2, so the endpoint is (-4, 2).
Make the radicand 0, 1, 4, 9:
| x + 4 | x | \displaystyle \sqrt{x + 4} | g(x) |
|---|---|---|---|
| 0 | -4 | 0 | 2 |
| 1 | -3 | 1 | 1 |
| 4 | 0 | 2 | 0 |
| 9 | 5 | 3 | -1 |
The minus sign in front of the root means each output is 2 minus something that grows, so the curve runs down from the endpoint; +x inside means it runs right.
The range is y \le 2, or (-\infty, 2]: the endpoint is now the highest point.
The table already shows both crossings. At x = 0 the output is 0, so the graph passes through the origin, and (0, 0) is both the y-intercept and the x-intercept. There is no other x-intercept, because the curve keeps falling: solving \displaystyle -\sqrt{x + 4} + 2 = 0 gives \displaystyle \sqrt{x + 4} = 2, so x + 4 = 4 and x = 0 only.
A curve that runs left
Graph \displaystyle h(x) = \sqrt{1 - x} + 2 and state its domain and range.
SolutionThe radicand 1 - x must be at least 0:
(add x to both sides). The domain is x \le 1, or (-\infty, 1]. At x = 1 the radicand is 0 and h(1) = 2, so the endpoint is (1, 2).
To make the radicand 0, 1, 4, 9, solve 1 - x = 0, 1, 4, 9, which gives x = 1, 0, -3, -8. The points are (1, 2), (0, 3), (-3, 4) and (-8, 5). Check one: \displaystyle h(-8) = \sqrt{9} + 2 = 5.
The inside has -x, so the inputs that work lie to the left of the endpoint and the curve runs left. The sign in front is +, so it runs up. The range is y \ge 2, or [2, \infty).
This graph is the mirror image, in the vertical line x = 1, of the graph of \displaystyle \sqrt{x - 1} + 2: the radicand at 1 - d equals the radicand of that other rule at 1 + d, since both are d.
A number in front of x inside the root
Graph \displaystyle p(x) = \sqrt{4x + 12} and state its domain and range. Then show that \displaystyle p(x) = 2\sqrt{x + 3}.
SolutionThe domain comes from the radicand: 4x + 12 \ge 0 gives 4x \ge -12, so x \ge -3, or [-3, \infty). At x = -3 the radicand is 0, so the endpoint is (-3, 0).
Making the radicand 1 would need 4x + 12 = 1, so \displaystyle x = -\frac{11}{4}, a fraction. Easier squares here are 0, 4, 16 and 36, all multiples of 4:
| 4x + 12 | x | p(x) |
|---|---|---|
| 0 | -3 | 0 |
| 4 | -2 | 2 |
| 16 | 1 | 4 |
| 36 | 6 | 6 |
The curve runs right (the inside has +4x) and up (the sign in front is +), so the range is y \ge 0, or [0, \infty).
For the second part, factor the radicand and split the root, which is allowed because 4 and x + 3 are both non-negative on the domain:
So p is the parent shifted left 3 and stretched vertically by 2, which is quicker to graph from: (1, 1), (4, 2), (9, 3) go to (-2, 2), (1, 4), (6, 6), the same points as the table. Squeezing the parent horizontally by 4 and stretching it vertically by 2 give the same curve. A trade like this works for powers of x such as x^2, with different numbers, but not for most functions: the graph of 2x + 1 is not a vertical stretch of the graph of x + 1.
Writing the rule from a graph
Find a rule of the form \displaystyle y = a\sqrt{x - h} + k for the graph below.
The endpoint is (-1, -2), so h = -1 and k = -2, and the rule so far is
The curve runs right, which fits +x inside, so this form is the right one. One more point fixes a. The point (3, 2) is on the graph, so
and a = 2. The rule is \displaystyle y = 2\sqrt{x + 1} - 2.
Check with the third marked point, (8, 4): \displaystyle 2\sqrt{9} - 2 = 6 - 2 = 4. It fits. A check with a different point matters here, because one point alone will produce some value of a for any curve with this endpoint; only a second point confirms that the curve really has this shape.
The answer is a rule, not the rule: the same graph is \displaystyle y = \sqrt{4x + 4} - 2, by the move in Example 4 run backwards. The form \displaystyle a\sqrt{x - h} + k was asked for because it has a single answer.
Reading and solving with the graph
(a) For the lead rule \displaystyle f(x) = 2\sqrt{x - 3} + 1, find the input with f(x) = 6, and say whether any input gives f(x) = 0.
(b) For \displaystyle g(x) = -\sqrt{x + 4} + 2 from Example 2, is there an input with g(x) = 3?
Solution(a) The graph rises through (7, 5) and (12, 7), so the height 6 is reached once, somewhere between x = 7 and x = 12. To find it exactly, undo the rule one step at a time:
Squaring in the fourth line is safe here, because \displaystyle \frac{5}{2} is not negative and so it really is the square root of \displaystyle \frac{25}{4}. Check: \displaystyle \frac{37}{4} - 3 = \frac{25}{4}, and \displaystyle 2 \cdot \frac{5}{2} + 1 = 6. As a decimal, x = 9.25, between 7 and 12 as the graph said.
No input gives f(x) = 0. The range is y \ge 1, and 0 is not in it: the whole curve sits at height 1 or above.
(b) No. The range of g is y \le 2, and 3 is above the endpoint. The algebra agrees, if read carefully: \displaystyle -\sqrt{x + 4} + 2 = 3 gives \displaystyle \sqrt{x + 4} = -1, and a square root is never negative. Squaring both sides anyway would give x + 4 = 1, so x = -3, but g(-3) = -1 + 2 = 1, not 3. The squaring step created a false answer. The graph shows at once why there is none.
Half of a sideways parabola
Show that the graph of \displaystyle f(x) = 2\sqrt{x - 3} + 1 is the upper half of the curve \displaystyle x = \frac{(y - 1)^2}{4} + 3, a parabola opening to the right.
SolutionTake a point (x, y) on the graph, so \displaystyle y = 2\sqrt{x - 3} + 1. Then
so every point of the graph lies on the parabola. It lies on the upper half only: the first line says y - 1 is twice a square root, so y - 1 \ge 0, that is y \ge 1.
In the other direction, take a point of the parabola with y \ge 1. Then (y - 1)^2 = 4(x - 3), and since y - 1 \ge 0, it is the non-negative square root of 4(x - 3), that is \displaystyle y - 1 = 2\sqrt{x - 3}. So the point is on the graph. Check with (7, 5): \displaystyle \frac{(5 - 1)^2}{4} + 3 = 4 + 3 = 7.
The lower half of the parabola, where y < 1, is the graph of \displaystyle -2\sqrt{x - 3} + 1. The whole parabola is not the graph of any function, since a vertical line such as x = 7 meets it at (7, 5) and (7, -3). Choosing the non-negative root is exactly what keeps one half and makes a function. This is the square root graph’s origin, too: it is half of a parabola y = x^2 turned on its side, and Restricting a domain to get an inverse runs the same idea in reverse.
Boundaries
The radicand must be at least 0, and only the radicand decides the domain. Numbers outside the root, such as the 2 and the +1 in \displaystyle 2\sqrt{x - 3} + 1, never change which inputs are allowed. They move the graph up, down and taller, never sideways.
Real outputs only. Over the complex numbers, \displaystyle \sqrt{-4} has a value, but a graph on the xy-plane plots only real outputs. That is why the graph stops at the endpoint.
The endpoint is (h, k) only in the form \displaystyle a\sqrt{x - h} + k. With a number in front of x inside, as in \displaystyle \sqrt{4x + 12}, the endpoint is where the radicand is 0, at x = -3, not at x = -12. Find it from the radicand, not by reading a number off the rule.
Pulling a number out of the root needs both factors non-negative. \displaystyle \sqrt{4(x + 3)} = 2\sqrt{x + 3} holds on the domain, where x + 3 \ge 0. The rule \displaystyle \sqrt{ab} = \sqrt{a}\sqrt{b} fails when both are negative: \displaystyle \sqrt{(-4)(-9)} = 6, but \displaystyle \sqrt{-4} and \displaystyle \sqrt{-9} are not real. Only a perfect square in front of x comes out as a whole number; \displaystyle \sqrt{2x} is \displaystyle \sqrt{2}\sqrt{x}.
These are rules with one square root of a first-degree radicand. A rule such as \displaystyle \sqrt{9 - x^2} has a radicand that is not of the form bx + c. Its domain is still found from the radicand, here -3 \le x \le 3, but its graph is a half circle with two ends, not a half curve with one, and the four directions do not apply.
Every output of the range is reached exactly once. Since the curve keeps rising (or keeps falling) from the endpoint, each height in the range is reached by one input and no more, and a height outside the range by none.
Squaring to solve can create false answers. When an equation such as \displaystyle \sqrt{x + 4} = -1 is squared, the result x + 4 = 1 has a solution the original does not. Check every answer in the original rule, or compare with the range first.
Where this goes next
The same endpoint, direction and range reasoning carries to every even root: \displaystyle \sqrt[4]{x} has the same half curve shape, a little flatter past (1, 1). Odd roots are different, and Cube root functions and roots as powers shows why the cube root accepts every input and runs through its center in both directions, with no endpoint at all.
The flattening shape is the reason square roots turn up in Radical functions in context: a pendulum four times as long swings only twice as slowly, and a car going twice as fast needs four times the braking distance, so speed grows like the square root of distance. Reading those situations is reading this graph.
Example 7 points at the larger idea. The parabola y = x^2 fails as an inverse because it repeats heights, and the square root keeps only one half. Restricting a domain to get an inverse makes that precise: cut x^2 down to x \ge 0 and its inverse is exactly \displaystyle \sqrt{x}, whose graph is the mirror image in the line y = x.
Exercises
Domain, endpoint and range
Find the domain, the endpoint and the range of each function, without graphing.
- \displaystyle y = \sqrt{x + 6} - 4
- \displaystyle y = 3\sqrt{x - 5}
- \displaystyle y = -\sqrt{x - 2} - 1
- \displaystyle y = \sqrt{8 - x} + 3
- \displaystyle y = -2\sqrt{-x} + 5
- \displaystyle y = \sqrt{3x + 12}
Graph each function
Plot the endpoint and three points with whole-number outputs, say which way the curve runs, and give the domain and range.
- \displaystyle y = \sqrt{x + 1} + 2
- \displaystyle y = 3\sqrt{x - 4} - 2
- \displaystyle y = -\sqrt{x - 1} + 5
- \displaystyle y = \sqrt{-x} - 3
- \displaystyle y = \sqrt{9x} - 1
Rules from graphs
Write a rule for each graph described.
- Endpoint (-3, 4), running right, through (1, 0). Use the form \displaystyle y = a\sqrt{x - h} + k.
- Endpoint (4, -1), running left, through (-5, 2). Use the form \displaystyle y = a\sqrt{h - x} + k.
- Endpoint (0, 0), running right and up, through (4, 10). Write the rule in two ways: once as a number times \displaystyle \sqrt{x}, and once as a single square root.
Solving with the graph
- For \displaystyle y = \sqrt{x + 1} + 2 (Exercise 7), find the input whose output is 6.
- Find the intercepts of \displaystyle y = -\sqrt{x - 1} + 5 (Exercise 9), or say why one does not exist.
- Find the intercepts of \displaystyle y = \sqrt{x + 16} - 3.
- Ksenia solves \displaystyle 3\sqrt{x - 4} - 2 = -5 like this: \displaystyle 3\sqrt{x - 4} = -3, so \displaystyle \sqrt{x - 4} = -1; squaring, x - 4 = 1, so x = 5. Is she right? Use the range to explain.
Reasoning
- Explain why the endpoint of \displaystyle y = a\sqrt{x - h} + k is its lowest point when a > 0 and its highest point when a < 0.
- Show that \displaystyle \sqrt{c^2 x} = c\sqrt{x} for every c > 0 and x \ge 0, and say what it means for squeezing the graph of \displaystyle \sqrt{x} horizontally.
- Explain why the graph of \displaystyle y = \sqrt{x} needs a run of 2n + 1 to rise from height n to height n + 1, and why that makes it flatten without ever leveling off.
- Show that the graph of \displaystyle y = -\sqrt{x + 4} + 2 is the lower half of the curve x = (y - 2)^2 - 4.
- Explain why no horizontal line meets the graph of \displaystyle y = a\sqrt{x - h} + k (with a \ne 0) in two points.
Answers
- Domain x \ge -6, that is [-6, \infty); endpoint (-6, -4); range y \ge -4, that is [-4, \infty). The radicand x + 6 is 0 at x = -6, and the sign in front is +.
- Domain [5, \infty); endpoint (5, 0); range [0, \infty). Nothing is added outside the root, so the endpoint sits on the x-axis.
- Domain [2, \infty); endpoint (2, -1); range (-\infty, -1]. The minus in front turns the curve down, so the endpoint is the highest point.
- Domain (-\infty, 8]; endpoint (8, 3); range [3, \infty). 8 - x \ge 0 gives x \le 8, and the curve runs left and up.
- Domain (-\infty, 0]; endpoint (0, 5); range (-\infty, 5]. -x \ge 0 gives x \le 0; the -2 in front sends the curve down.
- Domain [-4, \infty); endpoint (-4, 0); range [0, \infty). 3x + 12 \ge 0 gives x \ge -4.
- Points (-1, 2), (0, 3), (3, 4), (8, 5), from radicands 0, 1, 4, 9. Runs right and up. Domain [-1, \infty), range [2, \infty).
- Points (4, -2), (5, 1), (8, 4), (13, 7). Runs right and up, each rise 3 times the parent’s. Domain [4, \infty), range [-2, \infty).
- Points (1, 5), (2, 4), (5, 3), (10, 2). Runs right and down. Domain [1, \infty), range (-\infty, 5].
- Points (0, -3), (-1, -2), (-4, -1), (-9, 0), from -x = 0, 1, 4, 9. Runs left and up. Domain (-\infty, 0], range [-3, \infty).
- Points (0, -1), (1, 2), (4, 5), (9, 8), since \displaystyle \sqrt{9x} = 3\sqrt{x}. Runs right and up. Domain [0, \infty), range [-1, \infty).
- \displaystyle y = -2\sqrt{x + 3} + 4. From \displaystyle y = a\sqrt{x + 3} + 4 and the point (1, 0): \displaystyle 0 = a\sqrt{4} + 4, so 2a = -4 and a = -2.
- \displaystyle y = \sqrt{4 - x} - 1. From \displaystyle y = a\sqrt{4 - x} - 1 and (-5, 2): \displaystyle 2 = a\sqrt{9} - 1, so 3a = 3 and a = 1.
- \displaystyle y = 5\sqrt{x}, or \displaystyle y = \sqrt{25x}. From \displaystyle 10 = a\sqrt{4} = 2a, a = 5, and \displaystyle 5\sqrt{x} = \sqrt{25}\sqrt{x} = \sqrt{25x}.
- x = 15. \displaystyle \sqrt{x + 1} + 2 = 6 gives \displaystyle \sqrt{x + 1} = 4, so x + 1 = 16. Check: \displaystyle \sqrt{16} + 2 = 6.
- x-intercept (26, 0); no y-intercept. \displaystyle -\sqrt{x - 1} + 5 = 0 gives \displaystyle \sqrt{x - 1} = 5, so x = 26. The input 0 is not in the domain x \ge 1, so the graph never reaches the y-axis.
- y-intercept (0, 1), x-intercept (-7, 0). At x = 0, \displaystyle \sqrt{16} - 3 = 1. Setting \displaystyle \sqrt{x + 16} - 3 = 0 gives \displaystyle \sqrt{x + 16} = 3, so x + 16 = 9 and x = -7.
- No; the equation has no solution. The range of \displaystyle 3\sqrt{x - 4} - 2 is y \ge -2, and -5 is below it. Her step \displaystyle \sqrt{x - 4} = -1 is already impossible, since a square root is never negative; squaring hid that. Check: at x = 5 the rule gives \displaystyle 3\sqrt{1} - 2 = 1, not -5.
- \displaystyle \sqrt{x - h} is 0 at x = h and positive everywhere else in the domain. If a > 0 then \displaystyle a\sqrt{x - h} \ge 0, so y \ge k, with equality only at the endpoint: the endpoint is lowest. If a < 0 then \displaystyle a\sqrt{x - h} \le 0, so y \le k, and the endpoint is highest.
- Both \displaystyle c\sqrt{x} and \displaystyle \sqrt{c^2 x} are non-negative, and \displaystyle (c\sqrt{x})^2 = c^2 x. So \displaystyle c\sqrt{x} is the non-negative number whose square is c^2 x, which is what \displaystyle \sqrt{c^2 x} means. So squeezing the graph of \displaystyle \sqrt{x} horizontally by c^2 draws the same curve as stretching it vertically by c.
- Height n is reached at x = n^2 and height n + 1 at x = (n + 1)^2, since \displaystyle \sqrt{x} rises and each height is reached once. The run is (n + 1)^2 - n^2 = 2n + 1, which grows with n, so equal rises need longer runs and the curve flattens. It never levels off, because every height n, however large, is still reached, at x = n^2.
- From \displaystyle y = -\sqrt{x + 4} + 2, \displaystyle y - 2 = -\sqrt{x + 4}, and squaring gives (y - 2)^2 = x + 4, so x = (y - 2)^2 - 4. Since y - 2 is minus a square root, y - 2 \le 0, that is y \le 2: the lower half. Conversely, a point of that curve with y \le 2 has 2 - y \ge 0 and (2 - y)^2 = x + 4, so \displaystyle 2 - y = \sqrt{x + 4}, which is the rule.
- \displaystyle \sqrt{x - h} always rises: if h \le s < t then \displaystyle \sqrt{s - h} < \sqrt{t - h}, since squaring two non-negative numbers keeps their order. Multiplying by a > 0 keeps that order and by a < 0 reverses it, and adding k keeps it. So two different inputs never give the same output, and a horizontal line meets the graph at most once.