Quadratic functions: three forms and what each shows
One parabola, three costumes — and each one hands you a different feature for free.
The function f is defined by f(x) = x² − 2x − 8. Which of the following equivalent forms of f(x) displays the x-intercepts of the graph of y = f(x) in the xy-plane as constants or coefficients?
Factored form is the one that shows the zeros: (x − 4)(x + 2) = 0 at x = 4 and x = −2, so the x-intercepts are (4, 0) and (−2, 0), and the numbers 4 and −2 are sitting in the expression. A is equivalent, but it displays the vertex (1, −9), not the intercepts — the SAT loves that swap. C is not equivalent at all: it expands to x² + 2x − 8. D is equivalent but displays nothing; the −8 outside the product is not an intercept of anything.
1
One parabola, written three ways.
The three forms describe exactly the same curve. What changes is which feature you can read off without doing any work.
Same curve in all three panels: it crosses the y-axis at −8, crosses the x-axis
at −2 and 4, and bottoms out at (1, −9). The grey dots are the features that form
is hiding; the gold ones are the features it is handing you.
The word to watch is “displays”
A whole family of SAT items reads “Which of the following equivalent forms of
f(x) displays the … as constants or coefficients?” Every wrong choice
may be perfectly equivalent. The question is not which one is equal — it is
which one lets you read the thing off without calculating.
2
Standard form gives you the y-intercept.
f(x) = ax² + bx + c. Substitute x = 0 and the first two terms vanish, so c is the y-intercept — it is the only feature standard form shows for free.
f(x) = x² − 2x − 8
a = 1, b = −2, c = −8
f(0) = 0 − 0 − 8 = −8
y-intercept (0, −8)
a = 1 > 0
opens upward, so the vertex is a minimum
Standard form also carries the axis of symmetry, but only after one division:
x = −b2a = 2 ÷ 2 = 1. That is a calculation, not a read-off, which is why an
item asking for a form that displays the vertex will never accept
standard form.
3
Factored form gives you the x-intercepts.
f(x) = a(x − r)(x − s). A product is zero only when a factor is zero, so the zeros are r and s — and the zeros of f are exactly the x-intercepts of y = f(x).
f(x) = (x − 4)(x + 2)
from the opening problem
x − 4 = 0 → x = 4
first intercept (4, 0)
x + 2 = 0 → x = −2
second intercept (−2, 0)
The bracket shows the opposite sign
(x + 2) means the root −2, not 2. Read every bracket back to yourself by
setting it to zero before you answer — that one habit kills the most common
distractor in this whole topic, and it is why (x + 4)(x − 2) in the opening
problem was not even the same function.
Factored form hides the vertex, but not very well: the parabola is symmetric, so
the axis sits exactly midway between the two roots. Here (−2 + 4) ÷ 2 = 1, which
matches −b2a. Not every quadratic factors over the integers, though — when it
does not, the SAT gives it to you in another form or asks for something else.
4
Vertex form gives you the vertex.
f(x) = a(x − h)² + k has its vertex at (h, k) — h with its sign flipped, k exactly as written.
f(x) = (x − 1)² − 9
h = 1, k = −9
vertex (1, −9)
and the axis is x = 1
(x − 1)² ≥ 0 for every x
so f never drops below −9
minimum value of f = −9
reached at x = 1
vertex (1, −9), minimum value −9
“Minimum value of f” means the output −9. “The value of x at which f is least”
means the input 1. Vertex form displays both, one after the other, so read which
one the stem wants before you choose. Expand it if you doubt the equivalence:
(x − 1)² − 9 = x² − 2x + 1 − 9 = x² − 2x − 8 ✓
5
Converting between them is three moves, not nine.
From
To
Move
factored or vertex
standard
expand and collect
standard
factored
factor (find the pair)
standard
vertex
complete the square
factored
vertex
h is midway between the roots, then k = f(h)
That last row is the shortcut worth memorising. Straight from (x − 4)(x + 2), with no expanding at all:
h = (4 + (−2)) ÷ 2 = 1
midway between the roots
k = f(1) = (1 − 4)(1 + 2)
substitute into the factored form
k = (−3)(3) = −9
two small numbers, not a whole expansion
f(x) = (x − 1)² − 9
a stays whatever it was out front
The a never changes when you convert. If the factored form is 3(x − 4)(x + 2),
the vertex form is 3(x − 1)² + k with k = 3(−3)(3) = −27 — same h, and k computed
from the same substitution.
6
Going backwards: build the equation from its roots.
In the xy-plane, the graph of the quadratic function g has x-intercepts (−6, 0) and (2, 0), and passes through (0, 24). What is the y-coordinate of the vertex of the graph?
g(x) = a(x + 6)(x − 2)
a root of −6 gives the bracket (x + 6)
g(0) = a(6)(−2) = −12a
use the third point to pin a
−12a = 24, so a = −2
negative, so the parabola opens downward
h = (−6 + 2) ÷ 2 = −2
midway between the roots
g(−2) = −2(4)(−4) = 32
the vertex is (−2, 32), a maximum
g(x) = −2(x + 6)(x − 2), vertex (−2, 32)
Without a third point, a is not determined
Infinitely many parabolas share the same pair of x-intercepts — they differ in
how steep they are. Writing g(x) = (x + 6)(x − 2) assumes a = 1 for no reason,
and here it gives g(0) = −12 instead of 24. The same warning applies to the
vertex: a vertex plus a value of a, or a vertex plus one other point, but never
a vertex alone.
Expanded, g(x) = −2(x² + 4x − 12) = −2x² − 8x + 24 — and the 24 is the y-intercept
the stem gave you, which is a free check that you built the right function.
7
Match the stem to the form, then eliminate.
The stem asks for a form that displays…
Answer in
the y-intercept of the graph
standard form
the x-intercepts / the zeros / the solutions
factored form
the coordinates of the vertex
vertex form
the minimum (or maximum) value of the function
vertex form
the value of x where f is least
vertex form, then read h
zerosrootssolutionsall the same thing
Check equivalence before you check the feature
One choice is usually not equivalent at all, and it is the fastest one to throw
out: substitute x = 0 into it and into the given function. In the opening
problem, f(0) = −8, and (0 + 4)(0 − 2) = −8 as well — so that trick alone does
not settle it, and you have to expand: (x + 4)(x − 2) = x² + 2x − 8, with +2x
instead of −2x. Try a second value, x = 1, and it separates immediately:
f(1) = −9 while (5)(−1) = −5.
Desmos is on your screen — use it
The Digital SAT gives you a graphing calculator the whole time. Type the given
function and a candidate form as two equations: if one curve hides the other
completely, they are equivalent. Desmos also marks the intercepts and the vertex
as grey dots you can tap, which answers “what is the y-coordinate of the vertex”
in seconds. It cannot pick between two equivalent forms, though — for that,
match the feature word in the stem to the table above.
c is the y-intercept · the brackets are the x-intercepts · (h, k) is the vertex
Equivalent is not the question — displayed is.
Now you try
Practice problem
The function f is defined by f(x) = 3(x − 5)(x + 1). Which of the following equivalent forms of f(x) displays the coordinates of the vertex of the graph of y = f(x) in the xy-plane as constants or coefficients?
The roots are 5 and −1, so h = (5 + (−1)) ÷ 2 = 2 and k = f(2) = 3(2 − 5)(2 + 1) = 3(−3)(3) = −27. Vertex form is 3(x − 2)² − 27, and the vertex is (2, −27). A is equivalent but standard form — it displays the y-intercept −15, not the vertex. C keeps −15 as k, borrowing the constant from standard form; expanded it gives 3x² − 12x − 3, a different function. D flips the sign of h, putting the vertex at (−2, −27) instead.
Hint: you do not need to expand anything. The axis of symmetry sits midway
between the two roots, and once you have that x, one substitution into the
factored form gives you k. Then check the sign inside the bracket by reading it
back.
