Same shape, new address. Inside the bracket moves the curve sideways — opposite the sign — outside moves it up or down, and a leading minus turns it upside down.
Read the graph
Which equation matches the solid curve?
The vertex moved from (0, 0) to (2, 1): right 2 and up 1. Right 2 goes into the bracket with the opposite sign, (x − 2), and up 1 is added outside: y = (x − 2)² + 1. B would shift LEFT 2 — the inside-the-bracket sign always points the wrong way.
vertex: (0, 0) → (2, 1)
right 2, up 1 — the whole move
shape: unchanged
same width, same direction — a pure shift
A shift moves every point of the curve by the same amount, so tracking one recognisable point — the vertex, a corner, an intercept — captures the whole transformation.
shifts RIGHT h
minus inside → move right
shifts LEFT h
plus inside → move left
Why backwards? (x − 2)² needs x = 2 to produce what x² produced at 0 — the input has to catch up, so the picture slides toward positive x.
The vertex moved right 2, and right feels like “plus”, so the hand writes y = (x + 2)² + 1. Test its vertex: the bracket is zero at x = −2, so the curve sits at (−2, 1) — two squares left of the y-axis, the mirror image of the graph. Always check: set the bracket to zero and the x you get must be the vertex you see.
f(x) + k → up k
every output raised by k
f(x) − k → down k
outside is honest — no sign reversal
y = (x − 2)² + 1
the up-1 read straight off the vertex
| Change | Where it acts | Effect |
|---|---|---|
| y = f(x − h) | inside | slide right h (sign reversed) |
| y = f(x) + k | outside | slide up k (sign as written) |
| y = −f(x) | in front | flip upside down over the x-axis |
Flip first, then shift, and any parabola question of this type falls: an upside-down curve with vertex (h, k) is y = −(x − h)² + k before you have touched your pen to paper.
An odd middle number. The graph of y = x² − 5x + 4 is y = x² moved how?
x² − 5x + 254 − 254 + 4
half of 5 is 52; (52)² = 254
= (x − 52)² − 94
4 = 164, and 164 − 254 = −94
vertex (52, −94): right 2½, down 2¼
minus inside means right
What is different: an odd b makes the shifts fractions. Keep the quarters exact; rounding gives a vertex the graph does not have.
The expanded form starts with a minus. Decode y = −x² + 6x − 5.
= −(x² − 6x) − 5
take the minus out of the x terms only
= −(x² − 6x + 9) + 9 − 5
the 9 inside counts as −9, so add 9 outside
= −(x − 3)² + 4
vertex (3, 4)
flip, right 3, up 4
check x = 0: −9 + 4 = −5 ✓
What is different: the number we add inside the bracket is multiplied by the minus in front, so it is balanced by adding outside, not subtracting.
Writing −(x − 3)² − 9 − 5 = −(x − 3)² − 14 puts the vertex at (3, −14). Test x = 0 in it: −23, not the −5 the equation gives.
A different parent. Where is the corner of y = |x + 1| − 4, starting from y = |x|?
corner (0, 0) → (−1, −4)
plus inside: left 1; minus outside: down 4
(1, 1) on y = |x| → (0, −3)
move a second point the same way
|0 + 1| − 4 = −3 ✓
the equation agrees
What is different: nothing in the rules changes for |x|, √x or any other graph — we track the corner instead of a vertex.
Worked: the graph of y = x² + 8x + 13 is y = x² moved how?
x² + 8x + 16 − 16 + 13
half of 8 is 4; add and take away 4² = 16
= (x + 4)² − 3
vertex form
vertex (−4, −3): left 4, down 3
plus inside means left
The other direction — the choices are in standard form. Worked: an upside-down parabola with vertex (−1, 5).
y = −(x + 1)² + 5
flip, left 1, up 5
= −(x² + 2x + 1) + 5
square the bracket first
= −x² − 2x + 4
the minus reaches all three terms
Writing −x² + 2x + 1 + 5 changes only the first sign and lands on a planted choice. Ten-second check: put x = 0 into your answer and into −(0 + 1)² + 5 = 4 — the y-intercepts must agree.
Shift questions come as a graph to match, a point to move (f(x + 2) − 1 sends (3, 4) to (1, 3)), a standard-form equation to decode, or a flip followed by a shift — and the order matters: flip first, then move. Track the vertex, write vertex form, and expand only if the choices demand it.
Read the graph
Which equation matches the solid curve?
Two moves: the curve is upside down — a leading minus — and its vertex sits 3 up: y = −x² + 3. B shifts sideways to x = 3 instead of up; the vertex clearly stayed on the y-axis. Check a point: at x = 1, −1 + 3 = 2, one square below the vertex — exactly what the picture shows.
