Build a quadratic function from conditions

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Difficulty: Intermediate
Estimated time: 32 minutes
Domains: Advanced Math
Techniques: Quadratic-constructionPoint-substitutionSymmetryQuadratic-regression

What you’ll learn

Turn roots into factors, and a vertex into a squared term.
Use one more point to find $a$ , the number that sets the parabola’s scale: how wide it is and which way it opens.
Use symmetry to find a matching root or a matching point.
Fit a quadratic in standard form with regression from a table when all you’re given is three points, with no root or vertex named.
Check that your equation meets every condition in the question.

Prerequisites

Use the three forms of a quadratic function0%
Choose standard, factored, or vertex form because it shows the feature the problem gives you.

Why this matters on the SAT

Let the conditions choose the equation

The SAT doesn’t always hand you the equation. Sometimes it describes the parabola instead: its roots, its vertex, a point or two, or its symmetry. Those facts tell you which form to start from. After that, there’s usually one number still unknown, and one extra point pins it down.

Solution to the example

A root at $-2$ means the factor $x+2$ , since that’s $0$ when $x=-2$ . The same goes for $6$ and $x-6$ :

f(x)=a(x+2)(x-6).

Now use the point $(0,12)$ . It says $f(0)=12$ :

\begin{aligned} 12&=a(0+2)(0-6)\\[1.4em] 12&=-12a\\[1.4em] a&=-1. \end{aligned}

f(x)=-(x+2)(x-6),

and the answer is C.

The roots gave you the factors, and the extra point set the size and sign of $a$ . Choice B shows why you need that point: it has the right roots, but it gives $f(0)=(2)(-6)=-12$ , not $12$ . It opens up when this parabola has to open down. Here’s the idea to hold on to: the features pick the form, and one more point finds $a$ .

SAT example

The graph of a quadratic function $f$ has $x$ -intercepts at $x=-2$ and $x=6$ . The graph also passes through the point $(0,12)$ . Which choice could define $f$ ?

A
$f(x)=-(x-2)(x+6)$
B
$f(x)=(x+2)(x-6)$
C
$f(x)=-(x+2)(x-6)$
D
$f(x)=-(x+2)(x+6)$

Turn a graph into conditions

The facts a question gives you about a parabola are its conditions, and each one can become an equation. A point $(u,v)$ on the graph means $f(u)=v$ : put in $u$ , get out $v$ . A root is a point whose output is $0$ . The vertex is the turning point, and the vertical line through it is the axis of symmetry.

Each feature points to a form: the roots to factored form, and the vertex to vertex form. The extra point

(0,-8)

is the one that finds

a

Here’s how that works for this graph. The roots $-2$ and $4$ give

f(x)=a(x+2)(x-4).

The extra point $(0,-8)$ says $f(0)=-8$ :

-8=a(2)(-4),

so $a=1$ , and the equation is $f(x)=(x+2)(x-4)$ . Now check the condition you didn’t use. The axis sits halfway between the roots, at $x=1$ , and $f(1)=(3)(-3)=-9$ . That matches the vertex $(1,-9)$ .

Every problem like this takes the same four moves:

\boxed{\text{list the conditions}\ \longrightarrow\ \text{pick a form, or use regression}\ \longrightarrow\ \text{find the missing numbers}\ \longrightarrow\ \text{check every condition}.}

Common mistake:

Using a point backward. In $(u,v)$ , the first number $u$ is the input and the second number $v$ is the output. So substitute $x=u$ and set the function equal to $v$ . Once you’ve solved, put the point back into your finished equation and check both numbers.

Count what’s still unknown

How you build the function depends on how much of it the question has already filled in.

Roots or a vertex fill in most of the equation. Two roots fill in the $r$ and the $s$ in $a(x-r)(x-s)$ , and a vertex fills in the $h$ and the $k$ in $a(x-h)^2+k$ . Either way, only $a$ is left. So one more point gives you one equation with one unknown, and you can solve it by hand in a line or two. The answer is exact, and you can see how each fact shapes the equation. That’s why you start from the matching form, not from $ax^2+bx+c$ with three unknowns.

Now say all you’re given is three points, and the question doesn’t say any of them is a root or the vertex. We’ll call these three unrelated points. They don’t fill in any form, so $a$ , $b$ and $c$ are all unknown, and finding them by hand means solving three equations at once. That’s slow, and quadratic regression in Desmos does it for you.

So once you’ve written your starting form, count what’s still unknown. Just $a$ ? Go by hand. All three? Use regression.

Here’s where to start for each kind of fact, and what to do after that.

Start here

If you know the roots ( $x$ -intercepts), like $-2$ and $6$ , start with factored form, $f(x)=a(x+2)(x-6)$ .
If you know the vertex, like $(2,-5)$ , start with vertex form, $f(x)=a(x-2)^2-5$ .
If all you have is three unrelated points, like $(-1,6)$ , $(0,1)$ and $(2,3)$ , put them in a Desmos table and run quadratic regression.
If the question tells you a coefficient directly, like $c=4$ , start with standard form, $f(x)=ax^2+bx+c$ .

Then

After factored or vertex form, plug in one more point that isn’t a root or the vertex, and solve for $a$ .
After regression, define $f$ with the letters $a$ , $b$ and $c$ that Desmos stored, not the rounded numbers on screen, and evaluate $f$ .
If you know the axis of symmetry, mirror a root or a point across it to find its partner.
At the end, check the facts you didn’t build with, like the vertex when you started from the roots.

A fact that’s already built into the form can’t find $a$ . Every $a(x-2)^2-5$ with $a\ne0$ has its vertex at $(2,-5)$ , so plugging in the vertex can’t tell you which $a$ it is.

Roots plus one point

Say the roots are $r$ and $s$ . Start with

f(x)=a(x-r)(x-s).

The roots give you the factors, but not $a$ . Lots of parabolas cross the $x$ -axis at the same two roots. Some open up and some open down, and some are narrow while others are wide. A point that isn’t a root tells you which one the question means.

Check your understanding:

A quadratic has roots $1$ and $5$ and passes through $(3,-8)$ . Write the equation that finds $a$ in $f(x)=a(x-1)(x-5)$ , then solve it.

Vertex plus one point

Say the vertex is $(h,k)$ . Start with

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

The vertex is already built in, so plug in a different point to find $a$ . Then check its sign against the graph: $a$ is positive when the parabola opens up and negative when it opens down.

Common mistake:

Plugging a root into factored form, or the vertex into vertex form, to find $a$ . It seems like any given point should work. But at those points the part with $a$ equals $0$ , so you end up with something like $0=0$ or $k=k$ , which says nothing about $a$ . Use a point that isn’t already built into the form.

Use symmetry to find missing points

The axis of symmetry splits a parabola into two mirror images. If the axis is $x=h$ , two inputs the same distance from $h$ give the same output:

f(h-d)=f(h+d).

That gives you two shortcuts:

If one root is $h-d$ , the other root is $h+d$ .
If $(h-d,v)$ is on the graph, so is $(h+d,v)$ .

If the letters make this feel abstract, think of it as counting. Find how far the point you know is from the axis, then count the same distance on the other side. A quick number line sketch helps.

For example, say the axis is $x=3$ and one root is $-1$ . That root is $4$ units left of the axis, so the other root is $4$ units right of it:

3+4=7.

So you can write

f(x)=a(x+1)(x-7).

If the graph also passes through $(0,-14)$ , then

-14=a(1)(-7),

so $a=2$ and $f(x)=2(x+1)(x-7)$ .

Check your understanding:

A parabola has axis of symmetry $x=5$ and passes through $(2,11)$ . What other point must be on the graph, and why?

Build from three unrelated points

Here’s regression at work. The table holds the three points, and Desmos finds all three standard-form coefficients at once. If you haven’t run a regression before, the typing is the only new part, and it’s the same five steps every time.

Say a quadratic passes through $(-1,6)$ , $(0,1)$ and $(2,3)$ .

Enter the inputs in the x_1 column and the outputs in the y_1 column.
Enter y_1~a*x_1^2+b*x_1+c. The ~ tells Desmos to find the $a$ , $b$ and $c$ that make this equation fit the table.
Desmos shows $a=2$ , $b=-3$ and $c=1$ . It also stores each full, unrounded value under its letter.
Define f(x)=a*x^2+b*x+c. Type the letters $a$ , $b$ and $c$ , not the numbers.
If the question asks for $f(4)$ , enter f(4).

The fitted function is

f(x)=2x^2-3x+1,

f(4)=2(4)^2-3(4)+1=21.

Last, check all three given points:

f(-1)=6,\qquad f(0)=1,\qquad f(2)=3.

In the calculator, the curve passes through every table point. When the question gives exact values, all three checks have to match exactly. A decimal that’s close isn’t enough.

Check your understanding:

After a quadratic regression reports $a$ , $b$ and $c$ , why define $f(x)=ax^2+bx+c$ before you plug in a new input?

Calculator loads as you approach

Try changing a number in the table. The regression finds new values of a, b and c, and f, f(4) and the checked points all update with them.

No calculator? Or do you need to work out the exact coefficients with algebra? Then substitute the three points into $f(x)=ax^2+bx+c$ and solve the three equations by hand. That works, but for three unrelated points it’s usually slower.

Common mistake:

Using quadratic regression when the question already gives roots or a vertex. Regression can feel like the more powerful tool, so it’s tempting to use it everywhere. But there, only $a$ is missing, and one exact substitution usually finds it. The table setup costs you time and buys you nothing, so save regression for three unrelated points.

Example: build from a vertex and a point

Worked example

A quadratic function $f$ has a vertex at $(2,-5)$ and passes through $(5,13)$ . Which choice gives an equation for $f$ ?

A
$f(x)=(x-2)^2-5$
B
$f(x)=2(x-2)^2-5$
C
$f(x)=2(x+2)^2-5$
D
$f(x)=2(x-2)^2+5$

Step 1

Start from the vertex

The vertex $(h,k)=(2,-5)$ points straight to vertex form:

f(x)=a(x-2)^2-5.

The vertex sets $h$ and $k$ , so only $a$ is left to find.

Step 2

Turn the extra point into an equation

The point $(5,13)$ means $f(5)=13$ . Plug in both numbers:

13=a(5-2)^2-5.

Now solve:

\begin{aligned} 13&=9a-5\\[1.4em] 18&=9a\\[1.4em] a&=2. \end{aligned}

Step 3

Write the finished function

Put $2$ in for $a$ :

f(x)=2(x-2)^2-5.

The answer is B.

Step 4

Check every condition

At $x=2$ the squared term is $0$ , so $f(2)=-5$ . That’s the vertex. At the extra input,

f(5)=2(5-2)^2-5=18-5=13.

The graph shows the same thing: an upward-opening parabola through both points.

Calculator loads as you approach

You found the exact value of a by substitution. Now use the graph to confirm both conditions.

Try it yourself:

Before you look back at the algebra, explain why the vertex alone can’t tell choices A and B apart. Then use the point $(5,13)$ to explain why B has the right value of $a$ .

Check your equation against every condition

Whether you built it by hand or with regression, finish the same way. When a graph helps, graph your equation and make sure that:

both roots sit on the $x$ -axis
the vertex is where it should be
every given point is on the curve
the parabola opens the way the sign of $a$ says it should

If something’s off, the graph tells you where to look. Right roots or the right vertex, but a point off the curve? Then $a$ is wrong, so redo that substitution. Opening the wrong way? Then $a$ has the wrong sign. The check takes a few seconds, and it catches a flipped sign before it costs you the question.

Practice problems

Your turn. Before you write anything, look at what each question gives you and count what’s still unknown.

Use roots and a point

Practice problem

A quadratic function $f$ has zeros at $x=-4$ and $x=2$ . The graph of $y=f(x)$ passes through $(0,16)$ . What is the value of $f(3)$ ?

Your answer

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Build the factored form by hand first. Then graph it if you want to check the roots and the point.

Use a vertex and a point

Practice problem

A quadratic function $g$ has a vertex at $(-3,5)$ and passes through the point $(-1,-3)$ . What is the value of $g(0)$ ?

Your answer

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Build it by hand in vertex form. The calculator is here if you want to check the vertex and both points.

Find a root from symmetry

Practice problem

The graph of a quadratic function $p$ has an axis of symmetry at $x=2$ and one $x$ -intercept at $(-3,0)$ . The graph also passes through $(0,-15)$ . The function can be written as

p(x)=a(x+3)(x-r),

where $a$ and $r$ are constants. What is the value of $a$ ?

Your answer

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Find the mirror root first, then solve for a. Graph your result to check the axis and both intercepts.

Fit a quadratic through three points

Practice problem

A quadratic function $q$ has a graph that passes through the points $(0,4)$ , $(1,3)$ , and $(3,7)$ . What is the value of $q(4)$ ?

Your answer

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Put the three points in a table and fit a quadratic regression.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Roots $r$ and $s$ point to $f(x)=a(x-r)(x-s)$ , and a vertex $(h,k)$ points to $f(x)=a(x-h)^2+k$ .
A point $(u,v)$ means $f(u)=v$ . To find $a$ , use a point the form doesn’t already hold.
Around the axis $x=h$ , inputs the same distance away have equal outputs, and roots come in mirror pairs.
Count what’s still unknown. Just $a$ ? Go by hand. All three, from three unrelated points? Use quadratic regression in a table.
After regression, define the function with the stored $a$ , $b$ and $c$ , then evaluate what the question asks for.
Finish by checking every condition, including every given point.

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