Algebra 1 · A.SSE.B.3 · F.IF.C.8

Reading a quadratic in its three forms

The same parabola can be written three ways. Which way answers the question you were actually asked?

You need

Factored formAlgebra 1

Where you are 3 in Quadratic functions

Functions13

Quadratic functions3

Maximum, minimum and range of a quadratic function Quadratic functions in context

Reading a quadratic in its three forms

When you reach for it

Most questions about a quadratic function ask about one feature of its graph. How high or low does it go? Where does it cross the x-axis? Where does it start, at x = 0? A quadratic can be written in three forms, and each one puts a different feature in plain sight while hiding the others.

Reach for this skill when you have a quadratic in one form and a question that another form answers, and when you must compare two quadratic functions that arrive in different dress: one as an equation, the other as a graph, a table or a sentence. The conversions themselves are taught elsewhere: converting to vertex form and factored form. This lesson is about choosing which one to do.

The procedure

What each form shows

Form	Written	Shows at once
standard	ax^2 + bx + c	y-intercept (0, c)
vertex	a(x - h)^2 + k	vertex (h, k), axis x = h, greatest or least value k
factored	a(x - p)(x - q)	zeros p and q, axis \displaystyle x = \frac{p + q}{2}

Each feature sits where a piece vanishes. At x = 0 every term with x in it is 0. The square in vertex form is never negative and is 0 only at x = h. A product is 0 exactly when one of its factors is.

In every form, a is the same number, because each form multiplies out to a single x^2 term, ax^2. Its sign gives the direction: a > 0 opens up, so the vertex is a least value, and a < 0 opens down, so the vertex is a greatest value.

Here is one function written all three ways.

\begin{aligned} f(x) &= -\tfrac{1}{2}x^2 + 2x + \tfrac{5}{2} \\ &= -\tfrac{1}{2}(x - 2)^2 + \tfrac{9}{2} \\ &= -\tfrac{1}{2}(x + 1)(x - 5) \end{aligned}

Figure 1One parabola, three forms. Vertex form shows the top point, standard form the point on the y-axis, factored form the two points on the x-axis.

Why the three are one parabola. Multiply each out:

\begin{aligned} -\tfrac{1}{2}(x - 2)^2 + \tfrac{9}{2} &= -\tfrac{1}{2}(x^2 - 4x + 4) + \tfrac{9}{2} \\ &= -\tfrac{1}{2}x^2 + 2x + \tfrac{5}{2} \end{aligned}

\begin{aligned} -\tfrac{1}{2}(x + 1)(x - 5) &= -\tfrac{1}{2}(x^2 - 4x - 5) \\ &= -\tfrac{1}{2}x^2 + 2x + \tfrac{5}{2} \end{aligned}

Each step is the distributive property, which never changes a value. So all three expressions give the same output for every input x: they are one function, and they draw one graph.

One substitution away

A feature that a form does not show may still be close.

The y-intercept from any form. Put x = 0. Standard form just makes it free.
The vertex from factored form. A parabola is symmetric about its axis, and the two zeros are mirror images, so the axis is halfway between them. Substitute that x to get the height.
The zeros from vertex form. Set a(x - h)^2 + k = 0, so \displaystyle (x - h)^2 = -\frac{k}{a}, and finish with the square root property.

Standard form is the one furthest from both the vertex and the zeros. From it, you complete the square (with one more step when a \ne 1) to reach vertex form, or factor to reach factored form.

Choosing the form to produce

Read the question for the feature it asks about, then aim for the form that shows that feature.

The question asks	The feature	Aim for
greatest or least value, highest point, range, when it peaks	the vertex	vertex form
where the graph meets the x-axis, when the height is 0, solve f(x) = 0	the zeros	factored form
the starting value, the value at x = 0	c	standard form, or put x = 0

When both the vertex and the zeros are wanted, try factoring first. If it works, the zeros come at once and the vertex is one substitution away. If it does not, complete the square: vertex form gives the vertex, and the square root property gives the zeros.

When there is no factored form

Factored form needs zeros, real numbers p and q with f(p) = 0. Vertex form tells you whether they exist. If a > 0 and k > 0, then a(x - h)^2 \ge 0 for every x, so f(x) \ge k > 0: the graph sits wholly above the x-axis, there are no zeros, and there is no factored form a(x - p)(x - q) with real p and q. The same holds when a < 0 and k < 0, with the graph wholly below. Every quadratic has a standard form and a vertex form; only those that meet the x-axis have a factored form. Reading this straight from standard form is the job of the discriminant.

Comparing functions given in different ways

To compare a feature of two functions, get that same feature from each one, then compare numbers. From an equation, read it or convert. From a graph, read the marked point. From a table, use the symmetry: two inputs with equal outputs sit at equal distances from the axis, so the axis is halfway between them; a zero is an input whose output is 0; the starting value is the output at x = 0. From a sentence, build the equation from the facts it gives.

Worked examples

Example 1

What each form shows, and what one substitution adds

\begin{aligned} p(x) &= -3(x + 1)^2 + 7 \\ q(x) &= 4(x - 1)(x + 7) \\ r(x) &= x^2 - 8x + 5 \end{aligned}

(a) Say whether each function has a greatest or a least value, and find it. (b) Give each y-intercept.

Solution

(a) The sign of a decides: p has a = -3, so it opens down and has a greatest value; q and r have a = 4 and a = 1, so they have least values.

p is in vertex form, so its greatest value is 7, at x = -1.

q is in factored form, with zeros 1 and -7. The axis is halfway between them, \displaystyle x = \frac{1 + (-7)}{2} = -3, and substituting gives q(-3) = 4(-4)(4) = -64. The least value is -64.

r is in standard form, which hides the vertex, so complete the square: half of -8 is -4, and r(x) = (x - 4)^2 - 16 + 5, which is (x - 4)^2 - 11. The least value is -11, at x = 4.

(b) r shows its y-intercept: (0, 5). For the others, put x = 0: p(0) = -3(1) + 7 = 4 and q(0) = 4(-1)(7) = -28.

Example 2

Choosing the form for the question

For h(x) = x^2 + 12x + 32, find (a) where the graph crosses the x-axis, and (b) the least value of h.

Solution

(a) The question asks for the zeros, so aim for factored form. Two numbers with product 32 and sum 12 are 4 and 8:

h(x) = (x + 4)(x + 8)

The graph crosses the x-axis at x = -4 and x = -8.

(b) The least value is the vertex height. Completing the square would work, but the factored form already in hand is closer. Its axis is halfway between the zeros, \displaystyle x = \frac{-4 + (-8)}{2} = -6, and h(-6) = (-2)(2) = -4. The least value is -4.

Check with vertex form: (x + 6)^2 - 4 = x^2 + 12x + 36 - 4, which is x^2 + 12x + 32. When a quadratic factors, factored form answers both questions.

Example 3

Two questions, two forms

For k(x) = -3x^2 + 6x + 9: (a) does the graph ever rise above y = 13? (b) Where does it cross the x-axis?

Solution

(a) “Rise above” asks for the greatest value, so aim for vertex form. Take -3 out of the x terms, then complete the square inside:

\begin{aligned} k(x) &= -3(x^2 - 2x) + 9 \\ &= -3\big((x - 1)^2 - 1\big) + 9 \\ &= -3(x - 1)^2 + 12 \end{aligned}

Since a < 0, the greatest value is 12, at x = 1. The graph never reaches 13, so no.

(b) For the zeros, take out -3 and factor what is left:

k(x) = -3(x^2 - 2x - 3) = -3(x - 3)(x + 1)

The graph crosses at x = 3 and x = -1. The two forms agree: halfway between 3 and -1 is 1, the axis from vertex form.

Example 4

When there is no factored form

Write each in factored form if it has one: (a) m(x) = x^2 + 8x + 19; (b) n(x) = x^2 + 8x + 11.

Solution

Neither factors over the integers: no pair of integers with product 19, or 11, has sum 8. Vertex form decides.

(a) m(x) = (x + 4)^2 - 16 + 19, which is (x + 4)^2 + 3. Here a = 1 > 0 and k = 3 > 0, so m(x) \ge 3 for every x. The graph never reaches the x-axis, m has no zeros, and so it has no factored form with real numbers.

(b) n(x) = (x + 4)^2 - 5. The least value is -5 < 0 and the graph opens up, so it crosses the x-axis twice. Solve (x + 4)^2 = 5: \displaystyle x = -4 \pm \sqrt{5}. Each zero gives a factor:

n(x) = (x + 4 - \sqrt{5})(x + 4 + \sqrt{5})

Check: this is a difference of squares, (x + 4)^2 - 5, which is vertex form again.

Example 5

An equation against a graph

f(x) = -2(x - 3)^2 + 18, and g is the function graphed below.

Figure 2The graph of g in Example 5.

(a) Which function has the greater maximum? (b) Which has the higher y-intercept? (c) Which graph crosses the x-axis farther to the right?

Solution

Get each feature from each representation, then compare.

(a) f is in vertex form with a < 0: maximum 18, at x = 3. The top of the graph of g is (-2, 9): maximum 9. f has the greater maximum.

(b) f(0) = -2(9) + 18 = 0. The graph of g crosses the y-axis at (0, 5). g starts higher.

(c) The zeros of f: -2(x - 3)^2 + 18 = 0 gives (x - 3)^2 = 9, so x - 3 = \pm 3 and x = 0 or x = 6. The graph of g crosses at -5 and 1. f reaches farther right, 6 against 1.

Each question had its own winner, which is why you compare feature by feature. As a check on the reading, the zeros -5 and 1 give g(x) = a(x + 5)(x - 1), and the point (0, 5) gives a(5)(-1) = 5, so a = -1. Then g(-2) = -(3)(-3) = 9, matching the top of the graph.

Example 6

Two water jets

At a garden show two jets of water arc over a path. For jet A, a table gives the height of the water, y meters, at a distance x meters from its nozzle:

x	y
0	0
1	2.5
2	4
3	4.5
4	4
5	2.5
6	0

Jet B follows \displaystyle y = -\frac{1}{2}(x - 3)^2 + 5. (a) Which jet rises higher? (b) Which lands first, nearer its nozzle? (c) Which nozzle is higher off the ground?

Solution

(a) In jet A’s table, equal heights pair up around x = 3: 4 at x = 2 and x = 4, 2.5 at 1 and 5, 0 at 0 and 6. The axis is therefore x = 3, and the top is 4.5 meters. Jet B is in vertex form, top 5 meters at x = 3. Jet B rises higher.

(b) Jet A’s water is at height 0 at x = 6. For jet B, solve \displaystyle -\frac{1}{2}(x - 3)^2 + 5 = 0: (x - 3)^2 = 10, so \displaystyle x = 3 \pm \sqrt{10}. The value \displaystyle 3 - \sqrt{10} is negative, behind the nozzle, so it is not part of the arc. Jet B lands at \displaystyle 3 + \sqrt{10} \approx 6.16 meters, so jet A lands first.

(c) At x = 0: jet A is at 0, and jet B is at \displaystyle -\frac{1}{2}(0 - 3)^2 + 5 = 0.5. Jet B’s nozzle is 0.5 meters up, on the rim of a low basin.

The table can be turned into an equation too. Its zeros 0 and 6 give y = a\,x(x - 6), and the top (3, 4.5) gives a(3)(-3) = 4.5, so \displaystyle a = -\frac{1}{2}. As a check, x = 1 gives \displaystyle -\frac{1}{2}(1)(-5) = 2.5, as in the table.

Boundaries

Factored form needs real zeros. With a > 0 and k > 0, or a < 0 and k < 0, there is no factored form using real numbers. When k = 0 the graph touches the x-axis once, and a(x - h)^2 is factored form and vertex form at once, with the zero repeated.

c is the y-intercept only in standard form. In vertex form, k is the vertex height, which is the y-intercept only when h = 0.

A table shows only the rows it has. Its largest entry is the maximum only if the axis falls on a listed input. A pair of equal outputs fixes the axis exactly, since the function is quadratic.

A graph gives exact values only at marked or gridline points. Elsewhere a reading is an estimate.

In context, the domain may cut the parabola. A zero behind a nozzle, or a negative time, is a solution of the equation and not part of the situation.

All three forms need a \ne 0. With a = 0 the function is linear, with no vertex.

Where this goes next

When a quadratic will not factor and its zeros are wanted, the quadratic formula, coming later in this unit, finds them in one step from standard form, and the discriminant says beforehand whether there are any. The vertex as a greatest or least value is developed in maximum and minimum, and choosing a form to answer a question about a real situation is the daily work of quadratic functions in context.

Exercises

What the form shows

For f(x) = 3(x - 4)^2 - 5, give the vertex, the axis, and the least or greatest value.
For g(x) = -(x + 8)(x - 4), give the zeros, the axis and the greatest value.
For h(x) = -5x^2 + 2x + 9, give the y-intercept and say whether h has a greatest or a least value.
For p(x) = 3(x + 1)^2 + 2, find the y-intercept and say whether the graph meets the x-axis.

Choose the form, then produce it

Where does the graph of y = x^2 - 3x - 28 cross the x-axis?
Find the least value of y = x^2 - 10x + 18 and where it occurs.
For y = -x^2 + 8x - 12, find the greatest value and the zeros.
Find the range of y = 3x^2 - 12x + 4.
For q(x) = (x - 7)(x + 1), find the y-intercept and the vertex.

Factored form or not

Which of these have a factored form using real numbers? Give it where it exists. (a) (x + 1)^2 - 9 (b) -2(x - 3)^2 - 4 (c) 5(x - 2)^2
For which values of k does y = (x - 1)^2 + k have a factored form using real numbers?

Comparing across representations

Exercise 12 uses this table of a quadratic function g.

x	g(x)
-1	-5
0	0
1	3
2	4
3	3
4	0
5	-5

f(x) = -(x - 1)^2 + 7, and g is the function in the table. Which has the greater maximum? Which has the greater positive zero?
f(x) = 2(x - 3)(x - 9), and g is a quadratic whose graph has vertex (4, -20) and passes through (0, 12). Which has the smaller minimum? Which has the higher y-intercept?
f(x) = -(x - 4)^2 + 6, and g is a quadratic whose graph crosses the x-axis at 3 and 7 and the y-axis at -42. Which has the greater maximum?
A third jet at the show from Example 6 follows \displaystyle y = -\frac{1}{4}x(x - 7). Compared with jet B, which rises higher, and which lands farther from its nozzle?

Reasoning

Using factored form, show that f(x) = a(x - p)(x - q) takes the same value at x and at p + q - x, and explain why this puts the axis at \displaystyle x = \frac{p + q}{2}.
Both y = (x - 2)^2 + 1 and y = (x - 1)(x - 3) have a = 1 and axis x = 2. Are they the same function?
Explain why a quadratic with a > 0 and k > 0 in vertex form has no factored form using real numbers.

Answers

Vertex (4, -5), axis x = 4; a = 3 > 0, so least value -5.
Zeros -8 and 4; axis \displaystyle x = \frac{-8 + 4}{2} = -2; g(-2) = -(6)(-6) = 36, the greatest value since a < 0.
y-intercept (0, 9); a = -5 < 0, so a greatest value.
p(0) = 3(1) + 2 = 5, so (0, 5). Least value 2 > 0, so the graph never meets the x-axis.
At x = 7 and x = -4. Zeros, so factor: (x - 7)(x + 4).
Least value -7, at x = 5. Vertex form: (x - 5)^2 - 25 + 18, or (x - 5)^2 - 7.
Greatest value 4, at x = 4; zeros 2 and 6. -(x^2 - 8x) - 12 = -(x - 4)^2 + 4, and -(x^2 - 8x + 12) is -(x - 2)(x - 6).
y \ge -8. 3(x^2 - 4x) + 4 = 3(x - 2)^2 - 12 + 4, or 3(x - 2)^2 - 8, and a > 0.
q(0) = (-7)(1) = -7, so (0, -7). Axis halfway between 7 and -1 is x = 3, and q(3) = (-4)(4) = -16: vertex (3, -16).
(a) Yes: (x + 1)^2 = 9 gives zeros 2 and -4, so (x - 2)(x + 4). (b) No: a < 0 and k = -4 < 0, so the graph stays below the x-axis. (c) Yes: 5(x - 2)(x - 2), one repeated zero.
k \le 0. The least value is k; the graph meets the x-axis exactly when k \le 0.
f has the greater maximum, 7 against 4 (the table is symmetric about x = 2, where g = 4). g has the greater positive zero: g(4) = 0, while (x - 1)^2 = 7 gives f’s positive zero \displaystyle 1 + \sqrt{7} \approx 3.65.
g has the smaller minimum: f’s axis is x = 6 and f(6) = 2(3)(-3) = -18, against -20. f starts higher: f(0) = 2(-3)(-9) = 54, against 12.
g. From g(x) = a(x - 3)(x - 7) and g(0) = 21a = -42, a = -2. Its axis is x = 5 and g(5) = -2(2)(-2) = 8, greater than f’s 6.
Jet B rises higher: jet C’s axis is x = 3.5 and its top is \displaystyle -\frac{1}{4}(3.5)(-3.5), or \displaystyle \frac{49}{16} \approx 3.06, against 5. Jet C lands farther, at 7 against \displaystyle 3 + \sqrt{10} \approx 6.16.
f(p + q - x) = a(q - x)(p - x), and this is a(x - p)(x - q) = f(x), since (q - x)(p - x) is (-1)(x - q)(-1)(x - p). The inputs x and p + q - x have average \displaystyle \frac{p + q}{2}, so every height is shared by two points mirrored across \displaystyle x = \frac{p + q}{2}, which is therefore the axis.
No. (x - 2)^2 + 1 = x^2 - 4x + 5, but (x - 1)(x - 3) = x^2 - 4x + 3. Same a and axis, different vertices, (2, 1) and (2, -1): the first is 2 higher at every x.
(x - h)^2 \ge 0 and a > 0, so a(x - h)^2 + k \ge k > 0 for every x. No input gives 0, so there is no zero p; a factored form a(x - p)(x - q) would be 0 at x = p.