Use the three forms of a quadratic function

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Lesson progressPractice problems 0/3

Difficulty: Advanced
Estimated time: 27 minutes
Domains: Advanced Math
Techniques: Quadratic-formsEquivalent-expressionsInterceptsVertexAxis-of-symmetry

What you’ll learn

Pick standard, factored or vertex form based on what the question asks for.
Read the $y$ -intercept, zeros, vertex and axis of symmetry from the right form.
Rewrite one form as another without changing the function.
Get vertex form from standard form by completing the square, or from factored form by using the zeros and their midpoint.

Prerequisites

Why this matters on the SAT

Make the feature you need visible

The SAT will sometimes give you a quadratic in a form that hides the one thing the question asks about. You don’t need to work out everything about the parabola. You need the form that puts that feature in plain sight, so you either pick it from the choices or rewrite into it.

SAT example

The function $f$ is defined by

f(x)=2x^2-16x+23.

Which choice gives an equivalent form of $f$ that shows the minimum value of $f$ as a constant?

A
$f(x)=2(x-4)^2-9$
B
$f(x)=2(x-4)^2+23$
C
$f(x)=2(x-2)(x-6)-1$
D
$f(x)=2(x-8)^2+23$

Solution to the example

Every answer choice is a whole function, so you can let the graph check them for you. This is graph overlap:

Enter the original as y=2x^2-16x+23.
Enter choice A on a second line.
Does A sit right on top of the original everywhere? Then it’s the same function.
Does A show the feature the question asks for? Then you’re done.

Choice A covers the original completely. It’s also in vertex form, $a(x-h)^2+k$ , with a positive leading coefficient, so its constant $-9$ is the minimum value. A passes both checks, so the answer is A, and you don’t need to test the other choices.

Why check both? Look at choice C. It overlaps too, because $2(x-2)(x-6)-1$ multiplies out to $2x^2-16x+23$ . But its $-1$ isn’t the minimum. So when more than one choice overlaps, use a hybrid finish: the graph tells you which ones are the same function, and the question tells you which of those to pick.

Be careful with step 3, though. Choice D crosses the original at $(8,23)$ , but that’s the only point they share, so D is a different function. One shared point isn’t proof. The whole curve has to match.

When you have to write vertex form yourself

If there are no choices to test and the question asks you to write the exact vertex form from standard form, complete the square. To make a square from $x^2-8x$ , take half of $-8$ and square it:

\left(\frac{-8}{2}\right)^2=(-4)^2=16.

The $2$ outside the parentheses doubles that $16$ , so adding it really adds $32$ to the function. Subtract $32$ to balance it:

\begin{aligned} f(x) &=2(x^2-8x)+23\\[1.4em] &=2(x^2-8x+16)-32+23\\[1.4em] &=2(x-4)^2-9. \end{aligned}

Since $2(x-4)^2\ge 0$ , the smallest the output can be is $-9$ . So you have two tools here: graph overlap is the quicker way to pick from these choices, and completing the square is how you write the exact form yourself.

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Here are the original and choice A. They trace one parabola, so A is the same function, and its -9 is the minimum. No other choice needs testing.

One parabola, three useful views

Here’s one function written three ways:

f(x)=x^2-4x-5=(x+1)(x-5)=(x-2)^2-9.

All three give the same parabola. What changes is which numbers you can see.

It’s the same parabola in all three panels. Standard form shows the

y

-intercept,

(0,-5)

. Factored form shows the zeros,

-1

and

5

. Vertex form shows the vertex,

(2,-9)

, and the axis

x=2

Common mistake:

Changing the function, not just its form. A rewrite has to give the same output for every input, not just for one handy value. If a rewrite looks right but you’re not sure, multiply it back out, factor it back, or check several key features.

The leading coefficient $a$ is the same number in all three forms. It decides whether the parabola opens up or down, and how narrow or wide it is. The other numbers get rearranged so that each form shows you something different.

Pick the form from what the question asks for

Form	General form	What it shows	Why
Standard	$f(x)=ax^2+bx+c$	$y$ -intercept $(0,c)$	$f(0)=c$ , since every term with an $x$ becomes $0$
Factored	$f(x)=a(x-r)(x-s)$	Zeros $r$ and $s$	A product is $0$ when either factor is $0$
Vertex	$f(x)=a(x-h)^2+k$	Vertex $(h,k)$ and axis $x=h$	At $x=h$ the square is $0$ , and inputs the same distance from $h$ give the same output

Watch the signs. The sign inside a factor or a square is the opposite of the input it gives you:

$x-r=0$ gives the zero $x=r$ .
$x+5=x-(-5)$ , so the zero is $x=-5$ .
$(x+3)^2=(x-(-3))^2$ , so the vertex has $x$ -coordinate $-3$ .

Check your understanding:

The function $g$ is $g(x)=-4(x+2)^2+7$ . Without graphing, what are its vertex and axis of symmetry?

Common mistake:

Calling $k$ a minimum without checking $a$ . The vertex output is a minimum when $a>0$ and a maximum when $a<0$ . In $g(x)=-4(x+2)^2+7$ , $a$ is negative, so the parabola opens downward and $7$ is a maximum, not a minimum.

Choose before you rewrite

Read the last sentence of the question before you do any algebra. It tells you which feature you need, and that tells you which form to aim for. Let the question pick the form.

Once you know the feature, pick the quickest way to a form that shows it as a number.

Use graph overlap when…

the answer choices are whole functions, like the four versions of $f$ in the SAT example.
multiplying out a choice like $2(x-2)(x-6)-1$ would take longer than typing the original once and trying the choices one at a time.
you can stop at the first choice that passes both checks, the way A did in the SAT example.

Use hand algebra when…

you have to write or justify the exact form yourself. Multiply out for standard form and factor for factored form. For vertex form, complete the square from standard form, or use the midpoint of the zeros and its output from factored form.
the form you have already shows the feature, or one quick substitution like $f(0)$ gets you there faster.
you type in the answer yourself, so there are no choices to match on the graph.

If you can see it, read it. If you can pick it and graphing is quicker, graph it. If you have to write it, do the algebra.

Common mistake:

Doing more work than the question needs, like testing the other three choices after A has already passed, or multiplying out $-4(x+2)^2+7$ when the question only wants its vertex. Try one choice at a time and stop at the first that passes both checks. If the form you have already shows the number, read it off and move on.

You can also find the axis of symmetry from the two zeros. If the zeros are $r$ and $s$ , the vertex sits halfway between them:

x=\frac{r+s}{2}.

That midpoint gives you the axis, but not the vertex’s output. If you need the $y$ -coordinate too, plug the midpoint into the function.

This is also the quicker way from factored form to vertex form. For $f(x)=a(x-r)(x-s)$ , find $h=\frac{r+s}{2}$ , work out $k=f(h)$ , and write

f(x)=a(x-h)^2+k.

Try it on the parabola from earlier. The zeros of $(x+1)(x-5)$ are $-1$ and $5$ , so $h=\frac{-1+5}{2}=2$ and $k=(2+1)(2-5)=-9$ . That gives $(x-2)^2-9$ , the same vertex form as before, without multiplying anything out. There’s no need to multiply out a factored form and then complete the square unless the problem asks for that work.

Common mistake:

Reading the inside sign as the answer. If your zero or vertex input came out with the wrong sign, set the inside expression equal to $0$ and solve. For $x+4=0$ , the input is $-4$ , not $4$ .

Rewrite only as far as the feature requires

Take the shortest exact path to the form you need.

To standard form: multiply out and combine

Start from factored form:

\begin{aligned} p(x)&=3(x-2)(x+1)\\[1.4em] &=3(x^2-x-2)\\[1.4em] &=3x^2-3x-6. \end{aligned}

Now $p(0)=-6$ , so the $y$ -intercept is $(0,-6)$ . But if the question asks only for $p(0)$ , skip the multiplying and plug $0$ straight into the factored form:

p(0)=3(-2)(1)=-6.

To factored form: factor the trinomial

For

q(x)=2x^2+2x-12,

first pull out the common factor:

q(x)=2(x^2+x-6)=2(x+3)(x-2).

The zeros are $-3$ and $2$ . That’s the same factoring you already know.

To vertex form: complete one square

Here’s the whole method, one step at a time, on

r(x)=3x^2-12x+7.

First, factor the leading coefficient, $3$ , out of the $x^2$ and $x$ terms, and leave the constant outside:

r(x)=3(x^2-4x)+7.

Next, take half of the $x$ -coefficient inside the parentheses and square it. Half of $-4$ is $-2$ , and $(-2)^2=4$ . Adding $4$ inside turns $x^2-4x$ into the perfect square $(x-2)^2$ .

Here’s the same catch you saw with $f$ . That $4$ sits inside parentheses that get multiplied by $3$ , so it adds $3(4)=12$ to the function, not $4$ . Subtract $12$ outside and the function stays the same:

\begin{aligned} r(x) &=3(x^2-4x+4)-12+7\\[1.4em] &=3(x-2)^2-5. \end{aligned}

The vertex is $(2,-5)$ , and the axis is $x=2$ .

So the whole move is: add the square of half the $x$ -coefficient inside, then subtract that amount times the leading coefficient outside. In general, adding $m$ inside $a(\cdots)$ changes the function by $am$ , so you subtract $am$ , not $m$ . The number outside multiplies whatever you add inside.

Common mistake:

Subtracting only the number you added inside, like $4$ instead of $12$ above. It’s easy to miss, because you’re working inside the parentheses and the $a$ is out of sight. To catch it, multiply your vertex form back out and make sure you get the original standard form. The wrong version, $3(x-2)^2+3$ , multiplies out to $3x^2-12x+15$ , not $3x^2-12x+7$ .

Check your understanding:

In $5(x^2+6x)+1=5(x^2+6x+9)+1-\square$ , what number goes in the box, and why?

Example: show the vertex by completing the square

Worked example

The function $q$ is defined by

q(x)=2x^2+12x+11.

Which choice gives an equivalent form of $q$ that shows the minimum value of $q$ as a constant?

A
$q(x)=2(x+3)^2+11$
B
$q(x)=2(x+3)^2-7$
C
$q(x)=2(x-3)^2-7$
D
$q(x)=(2x+3)^2+2$

Step 1

Use overlap to find the equivalent choice

The choices are whole functions, so the graph is the quicker first move. Graph y=2x^2+12x+11, then try one choice at a time on a second line. Only choice B lands right on top of the original, so the answer is B.

Step 2

If you had to write it yourself

Now suppose there were no choices, and you had to write the vertex form yourself. Complete the square, just as with $r(x)$ . Factor $2$ out of the two terms with $x$ :

q(x)=2(x^2+6x)+11.

Now the number you halve is $6$ , not $12$ .

Step 3

Make a perfect square, then balance it

Half of $6$ is $3$ , and $3^2=9$ . Add $9$ inside the parentheses. The $2$ outside turns that into $18$ , so subtract $18$ :

\begin{aligned} q(x) &=2(x^2+6x+9)+11-18\\[1.4em] &=2(x+3)^2-7. \end{aligned}

Step 4

Read the feature and check

The vertex is $(-3,-7)$ . The leading coefficient is positive, so the parabola opens upward and $-7$ is the minimum value. The answer is B.

Multiply it back out to check the rewrite:

2(x+3)^2-7 =2(x^2+6x+9)-7 =2x^2+12x+11.

Try it yourself:

Now that you can see the vertex form, explain why $(x+3)^2$ puts the axis at $x=-3$ , not $x=3$ . Then explain why multiplying back out to the original standard form checks every input, while checking one function value checks only one point.

Practice problems

Try these on your own. For each one, ask yourself: does the form I’m given already show the answer, can I test the choices on the graph, or do I have to write the form myself?

Read the zeros and the axis

Practice problem

The function $f$ is defined by

f(x)=-2(x-3)(x+5).

Which statement must be true?

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Read the factors first. Desmos is here if you want to confirm the intercepts and axis.

Pick the form that shows a minimum

Practice problem

The function $p$ is defined by

p(x)=3x^2-18x+20.

Which choice gives an equivalent form of $p$ that shows the minimum value of $p$ as a constant?

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Graph the original, then try choice A on a second line.

Put two forms together

Practice problem

The quadratic function $g$ is defined by

g(x)=a(x-2)(x-10),

where $a$ is a constant. If the $y$ -intercept of the graph of $y=g(x)$ is $(0,40)$ , the function can be written as

g(x)=a(x-6)^2+k,

where $k$ is a constant. What is the value of $k$ ?

Your answer

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Find

a

from the factored form first. Desmos can confirm the vertex afterward if that helps.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Standard form $ax^2+bx+c$ shows the $y$ -intercept $(0,c)$ .
Factored form $a(x-r)(x-s)$ shows the zeros $r$ and $s$ .
Vertex form $a(x-h)^2+k$ shows the vertex $(h,k)$ and the axis $x=h$ .
Let the question pick the form, then rewrite only as far as you need to.
When the choices are whole functions and graphing is quicker, stop at the first choice that passes both checks: it matches the whole curve, so it’s the same function, and it shows what the question asks for.
To write vertex form from standard form, complete the square, and subtract what you added times $a$ .
To write vertex form from factored form, find the midpoint of the zeros, evaluate the function there, and keep the same leading coefficient.
Equivalent forms share every function value and every point on the graph.

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Build a quadratic function from conditions

Use roots, a vertex, points, or symmetry conditions to build a quadratic function.

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