Absolute value is distance from zero, so every question about it splits into two — and the graph folds into a V.
The given equation has two solutions. What is the sum of the two solutions?
| 3x + 6 | = 15
The inside is either 15 or −15. From 3x + 6 = 15 you get 3x = 9 and x = 3; from 3x + 6 = −15 you get 3x = −21 and x = −7. The sum is 3 + (−7) = −4. D solves only the positive case and stops, the single most common slip in this topic. C assumes the two answers are opposites — they are mirror images of each other about the centre −2, not about 0. A divides the 15 by 3 first without dividing the 6 as well, getting |x + 6| = 5 and the pair −1, −11.
|−7| and |7| are both 7. Distance has no sign, so two different numbers can have the same absolute value.
That one sentence is the whole topic. It is why |x| = 7 has two answers, why the graph of y = |x| has two arms, and why an inequality turns into either one stretch of the number line or two.
Whatever is between the bars equals c, or it equals −c. Solve both, keep both.
3x + 6 = 15
x = 3
3x + 6 = −15
x = −7
3x = 9 → x = 3
− 6 both sides, then ÷ 3
3x = −21 → x = −7
−15 − 6 = −21, not −9
For 5|x − 2| − 3 = 17, undo the −3 and the 5 first: 5|x − 2| = 20, so |x − 2| = 4, and only now split into x − 2 = 4 and x − 2 = −4, giving x = 6 and x = −2. Splitting while the 5 and the 3 are still attached scatters them into the wrong places. The same carelessness about what a number is being applied to is what puts −12 in the opening list of choices: dividing the 15 by 3 without dividing the 6 as well turns the equation into |x + 6| = 5.
And |2x + 1| + 9 = 4 has no solution at all: it says |2x + 1| = −5, and a distance is never negative. The SAT does ask this, usually as “how many solutions does the given equation have”.
Outside the bars the function is an ordinary line; the bars fold everything below the axis back up.
y = |2x − 6| + 1
where is the inside zero?
2x − 6 = 0 → x = 3
the fold happens here
y = |0| + 1 = 1
put x = 3 back in
vertex (3, 1)
every other point is higher
Pulling the 2 out of the bars is legal because |2u| = 2|u|. It is worth doing: once the coefficient of x inside is 1, the vertex and the steepness can be read straight off.
Vertex (h, k). The a sets the steepness of both arms, and a negative a turns the V upside down.
| Letter | What it does | Careful |
|---|---|---|
| h | slides the V h units right | it is x − h, so |x + 4| means h = −4 |
| k | slides the V k units up | k is added outside the bars |
| |a| > 1 | narrower — the arms climb faster | slope +a on the right, −a on the left |
| 0 < |a| < 1 | wider — the arms climb slower | |
| a < 0 | opens downward; the vertex is the maximum | the V still has its point at (h, k) |
y = |x| has its point at (0, 0)
the dashed V
y = 2|x − 3| − 4 has its point at (3, −4)
3 right, 4 down
x-intercepts: |x − 3| = 2, so x = 1 and x = 5
set y = 0 and split, as in step 2
The two x-intercepts always sit the same distance either side of the vertex, so their average is h. An SAT item that asks for the sum of the x-intercepts is really asking for 2h — here 1 + 5 = 6 = 2(3), and you can answer it without finding either intercept.
Read |x − c| as “the distance from x to c”. Then the inequality describes a region around c.
|x − 5| ≤ 4
within 4 of the number 5
−4 ≤ x − 5 ≤ 4
one inequality, three parts
1 ≤ x ≤ 9
+ 5 across the whole thing
|2x − 7| > 3
more than 3 away from the centre
2x − 7 > 3 or 2x − 7 < −3
two separate statements
x > 5 or x < 2
+ 7, then ÷ 2, in each one
Writing 3 < 2x − 7 < −3 claims a number is above 3 and below −3 at the same time, which nothing satisfies. A “greater than” answer needs the word or; only “less than” may be written as a sandwich.
The Digital SAT keeps a graphing calculator on screen. Enter y = |2x − 7| and y = 3 and look at where the V sits above the flat line: left of x = 2 and right of x = 5. You can also type the inequality |2x − 7| > 3 directly and Desmos shades the solution on the x-axis.
Absolute value answers are cheap to verify — the bars turn every sign question into arithmetic.
In the given function, a is a constant. In the xy-plane, the graph of y = f(x) passes through the point (1, −2). What is the value of a?
f(x) = a| x − 4 | + 7
Put x = 1 and y = −2 in: a|1 − 4| + 7 = −2. Since |−3| = 3, that is 3a + 7 = −2, so 3a = −9 and a = −3. Check: −3|1 − 4| + 7 = −9 + 7 = −2 ✓ A stops at 3a = −9 and reports the −9; C swaps the coordinates, using x = −2 and y = 1 to get 6a + 7 = 1; D drops the absolute value and writes −3a + 7 = −2, which is the same arithmetic with the sign of the whole thing flipped. The vertex here is (4, 7) and a is negative, so this V opens downward.
Hint: “passes through (1, −2)” means substitute x = 1 and y = −2 and solve for the constant. Work out the bars completely before you touch a — |1 − 4| is 3, not −3.
