Constructing Quadratic Functions

Using Known Points or Roots to Determine a Quadratic

Constructing a quadratic function means determining $f(x) = ax^2 + bx + c$ from the given information. Known points lead to a system for the coefficients, known roots lead to factored form, and a known vertex leads to vertex form.

Forms of Quadratic Functions

Standard Form: $f(x) = ax^2 + bx + c$
Factored Form: $f(x) = a(x - p)(x - q)$ when the real roots $p$ and $q$ exist
Vertex Form: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex point

These three forms are interrelated and can be converted from one form to another.

Methods of Constructing Quadratic Functions

Which conditions are given decides which approach reaches the function most quickly. Three points, a vertex with one more point, and the two intercepts each supply exactly the information one of the three forms is built from.

A quadratic function is fixed once enough conditions are known.

Three Points

Three points with distinct $x$ -coordinates determine exactly one polynomial of degree at most $2$ . It is a quadratic function only when the resulting coefficient satisfies $a\neq0$ .

If we have three points $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ , we can substitute these values into the standard equation to get three linear equations with three variables $a$ , $b$ , and $c$ .

y_1 = ax_1^2 + bx_1 + c

Example:

Find the quadratic function that passes through points $K(-1, 0)$ , $L(0, -3)$ , and $M(1, -4)$ .

Substitute the coordinate values into the standard equation.

K(-1, 0): 0 = a(-1)^2 + b(-1) + c

Simplify these equations.

0 = a - b + c

Substitute $c = -3$ into the first and third equations.

0 = a - b - 3

Simplify and add the two equations to find the value of $a$ .

a - b = 3

Substitute the value of $a$ into the equation $a - b = 3$ to find the value of $b$ .

1 - b = 3

The resulting quadratic function is:

f(x) = x^2 - 2x - 3

Graph of

f(x) = x^2 - 2x - 3

Parabola passing through points

K(-1, 0)

L(0, -3)

, and

M(1, -4)

Vertex Point

If we know the coordinates of the vertex $(h, k)$ and another point $(p, q)$ on the graph, we can use the vertex form $f(x) = a(x - h)^2 + k$ to find the value of $a$ .

Example:

Find the quadratic function that has a vertex at $(2, 0)$ and passes through the point $(4, 4)$ .

Use the vertex form $f(x) = a(x - h)^2 + k$ with $h = 2$ and $k = 0$ .

f(x) = a(x - 2)^2 + 0 = a(x - 2)^2

Substitute the point $(4, 4)$ to find the value of $a$ .

4 = a(4 - 2)^2

The resulting quadratic function is:

f(x) = (x - 2)^2 = x^2 - 4x + 4

Graph of

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

Parabola with vertex at

(2, 0)

and passing through point

(4, 4)

Roots and One Additional Point

If we know the roots ( $x$ -intercepts) $p$ and $q$ of the quadratic function, and an additional point $(r, s)$ on the curve with $r\neq p$ , $r\neq q$ , and $s\neq0$ , we can use the factored form $f(x) = a(x - p)(x - q)$ to determine $a$ .

Example:

Find the quadratic function that has roots at $x = -2$ and $x = 3$ , and passes through the point $(1, -6)$ .

Use the factored form $f(x) = a(x - p)(x - q)$ with $p = -2$ and $q = 3$ .

\begin{aligned} f(x) &= a(x - (-2))(x - 3) \\ &= a(x + 2)(x - 3) \end{aligned}

Substitute the point $(1, -6)$ to find the value of $a$ .

-6 = a(1 + 2)(1 - 3)

The resulting quadratic function is:

\begin{aligned} f(x) &= (x + 2)(x - 3) \\ &= x^2 - x - 6 \end{aligned}

Graph of

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

Parabola with roots at

x = -2

and

x = 3

, passing through point

(1, -6)

Choosing a Function from Two Conditions

The axis of symmetry and the discriminant usually allow several quadratic functions. Another piece of information is needed to select one unique function.

Example:

Find a quadratic function with axis of symmetry $x = 1$ and discriminant $D = 16$ .

From the axis of symmetry $x = 1$ , we know that $-\frac{b}{2a} = 1$ , so $b = -2a$ .

From the discriminant $D = 16$ , we know that $b^2 - 4ac = 16$ .

Substitute $b = -2a$ into the discriminant equation.

(-2a)^2 - 4ac = 16

There are many values of $a$ and $c$ that satisfy this equation, so the data do not determine a unique function. To obtain one example, choose $a = 1$ .

1(1 - c) = 4

With $a = 1$ , $b = -2$ , and $c = -3$ , the resulting quadratic function is:

f(x) = x^2 - 2x - 3

Symmetric Coordinates

Symmetric points can also determine a quadratic function.

Example:

Determine the quadratic function that passes through the points $(0, 0)$ , $(4, 1)$ , and $(-4, 1)$ .

Since the points $(4, 1)$ and $(-4, 1)$ have the same $y$ -value and are at the same distance from the $y$ -axis, the curve is symmetric about the $y$ -axis. This means the axis of symmetry is $x = 0$ , and the vertex is at $(0, 0)$ .

Use the vertex form $f(x) = a(x - h)^2 + k$ with $h = 0$ and $k = 0$ .

f(x) = ax^2

Substitute the point $(4, 1)$ to find the value of $a$ .

1 = a(4)^2

The resulting quadratic function is:

f(x) = \frac{1}{16}x^2

Graph of

f(x) = \frac{1}{16}x^2

Parabola passing through points

(0, 0)

(4, 1)

, and

(-4, 1)

Transformations Between Quadratic Function Forms

Each conversion takes one algebraic step, and the graph stays the same throughout.

The same quadratic function can be written in standard, factored, or vertex form.

Standard Form to Vertex Form

This conversion needs completing the square, because that step produces the squared bracket $(x-h)^2$ . The working below shows every intermediate step.

To convert $f(x) = ax^2 + bx + c$ to vertex form $f(x) = a(x - h)^2 + k$ :

Determine the $x$ -coordinate of the vertex: $h = -\frac{b}{2a}$
Calculate the function value at the vertex: $k = f(h) = f(-\frac{b}{2a})$
Or use the formula: $k = c - \frac{b^2}{4a}$

Standard Form to Factored Form

To convert $f(x) = ax^2 + bx + c$ to factored form over the real numbers, first verify that the discriminant is nonnegative:

Determine the roots of the equation $ax^2 + bx + c = 0$ using the formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
If $D\ge0$ and the real roots are $p$ and $q$ , then:

$f(x) = a(x - p)(x - q)$

If $D<0$ , the function has no real roots and cannot be written as a product of real linear factors.

Factored Form to Standard Form

To convert $f(x) = a(x - p)(x - q)$ to standard form $f(x) = ax^2 + bx + c$ :

f(x) = a(x - p)(x - q)

By comparing with the standard form, we get:

$b = -a(p+q)$
$c = apq$

Exercises and Solutions

Each exercise gives a set of conditions and asks for the matching function. Build the function, then check that its graph passes through every given point.

Start each solution from the form that fits the given conditions, and fill in the values only after that choice is made. Compare your own approach step by step with the approach in the worked solution, so that a wrong final value still shows which step differs.

First Exercise

Find the quadratic function that passes through the points $(-2, 4)$ , $(1, -5)$ , and $(3, 7)$ .

Answer:

Substitute the points into the standard equation:

\begin{aligned} 4 &= a(-2)^2 + b(-2) + c \\ &= 4a - 2b + c \end{aligned}

From the second equation, $c = -5 - a - b$ . Substitute into the first equation, then solve:

4 = 4a - 2b + (-5 - a - b)

Substitute into the third equation:

\begin{aligned} 7 &= 9(3 + b) + 3b \\ &\quad{}+ (-5 - (3 + b) - b) \end{aligned}

Then:

a = 3 + (-\frac{6}{5}) = \frac{9}{5}

The quadratic function is:

f(x) = \frac{9}{5}x^2 - \frac{6}{5}x - \frac{28}{5}

Second Exercise

Determine the quadratic function that has a vertex at $(-1, 4)$ and passes through the point $(2, -5)$ .

Answer:

Use the vertex form with $h = -1$ and $k = 4$ :

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

Substitute the point $(2, -5)$ :

-5 = a(2 + 1)^2 + 4

The quadratic function is:

\begin{aligned} f(x) &= -(x + 1)^2 + 4 \\ &= -x^2 - 2x + 3 \end{aligned}

Third Exercise

Determine the quadratic function that has roots at $x = -3$ and $x = 2$ , and has a maximum value of $4$ .

Answer:

Use the factored form with $p = -3$ and $q = 2$ :

\begin{aligned} f(x) &= a(x - (-3))(x - 2) \\ &= a(x + 3)(x - 2) \end{aligned}

Since the function has a maximum value, $a < 0$ .

The vertex is at $x = \frac{p + q}{2} = \frac{-3 + 2}{2} = -\frac{1}{2}$ .

Substitute $x = -\frac{1}{2}$ into the factored form and use the fact that the maximum value is $4$ :

\begin{aligned} f(-\frac{1}{2}) &= a(-\frac{1}{2} + 3)(-\frac{1}{2} - 2) \\ &= -a\frac{25}{4} \\ &= 4 \end{aligned}

The quadratic function is:

\begin{aligned} f(x) &= -\frac{16}{25}(x + 3)(x - 2) \\ &= -\frac{16}{25}x^2 - \frac{16}{25}x + \frac{96}{25} \end{aligned}

Using Known Points or Roots to Determine a Quadratic

Forms of Quadratic Functions

Standard Form: $f(x) = ax^2 + bx + c$
Factored Form: $f(x) = a(x - p)(x - q)$ when the real roots $p$ and $q$ exist
Vertex Form: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex point

These three forms are interrelated and can be converted from one form to another.

Methods of Constructing Quadratic Functions

A quadratic function is fixed once enough conditions are known.

Three Points

Three points with distinct $x$ -coordinates determine exactly one polynomial of degree at most $2$ . It is a quadratic function only when the resulting coefficient satisfies $a\neq0$ .

y_1 = ax_1^2 + bx_1 + c

Example:

Find the quadratic function that passes through points $K(-1, 0)$ , $L(0, -3)$ , and $M(1, -4)$ .

Substitute the coordinate values into the standard equation.

K(-1, 0): 0 = a(-1)^2 + b(-1) + c

Simplify these equations.

0 = a - b + c

Substitute $c = -3$ into the first and third equations.

0 = a - b - 3

Simplify and add the two equations to find the value of $a$ .

a - b = 3

Substitute the value of $a$ into the equation $a - b = 3$ to find the value of $b$ .

1 - b = 3

The resulting quadratic function is:

f(x) = x^2 - 2x - 3

Graph of

f(x) = x^2 - 2x - 3

Parabola passing through points

K(-1, 0)

L(0, -3)

, and

M(1, -4)

Vertex Point

If we know the coordinates of the vertex $(h, k)$ and another point $(p, q)$ on the graph, we can use the vertex form $f(x) = a(x - h)^2 + k$ to find the value of $a$ .

Example:

Find the quadratic function that has a vertex at $(2, 0)$ and passes through the point $(4, 4)$ .

Use the vertex form $f(x) = a(x - h)^2 + k$ with $h = 2$ and $k = 0$ .

f(x) = a(x - 2)^2 + 0 = a(x - 2)^2

Substitute the point $(4, 4)$ to find the value of $a$ .

4 = a(4 - 2)^2

The resulting quadratic function is:

f(x) = (x - 2)^2 = x^2 - 4x + 4

Graph of

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

Parabola with vertex at

(2, 0)

and passing through point

(4, 4)

Roots and One Additional Point

Example:

Find the quadratic function that has roots at $x = -2$ and $x = 3$ , and passes through the point $(1, -6)$ .

Use the factored form $f(x) = a(x - p)(x - q)$ with $p = -2$ and $q = 3$ .

\begin{aligned} f(x) &= a(x - (-2))(x - 3) \\ &= a(x + 2)(x - 3) \end{aligned}

Substitute the point $(1, -6)$ to find the value of $a$ .

-6 = a(1 + 2)(1 - 3)

The resulting quadratic function is:

\begin{aligned} f(x) &= (x + 2)(x - 3) \\ &= x^2 - x - 6 \end{aligned}

Graph of

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

Parabola with roots at

x = -2

and

x = 3

, passing through point

(1, -6)

Choosing a Function from Two Conditions

The axis of symmetry and the discriminant usually allow several quadratic functions. Another piece of information is needed to select one unique function.

Example:

Find a quadratic function with axis of symmetry $x = 1$ and discriminant $D = 16$ .

From the axis of symmetry $x = 1$ , we know that $-\frac{b}{2a} = 1$ , so $b = -2a$ .

From the discriminant $D = 16$ , we know that $b^2 - 4ac = 16$ .

Substitute $b = -2a$ into the discriminant equation.

(-2a)^2 - 4ac = 16

There are many values of $a$ and $c$ that satisfy this equation, so the data do not determine a unique function. To obtain one example, choose $a = 1$ .

1(1 - c) = 4

With $a = 1$ , $b = -2$ , and $c = -3$ , the resulting quadratic function is:

f(x) = x^2 - 2x - 3

Symmetric Coordinates

Symmetric points can also determine a quadratic function.

Example:

Determine the quadratic function that passes through the points $(0, 0)$ , $(4, 1)$ , and $(-4, 1)$ .

Use the vertex form $f(x) = a(x - h)^2 + k$ with $h = 0$ and $k = 0$ .

f(x) = ax^2

Substitute the point $(4, 1)$ to find the value of $a$ .

1 = a(4)^2

The resulting quadratic function is:

f(x) = \frac{1}{16}x^2

Graph of

f(x) = \frac{1}{16}x^2

Parabola passing through points

(0, 0)

(4, 1)

, and

(-4, 1)

Transformations Between Quadratic Function Forms

Each conversion takes one algebraic step, and the graph stays the same throughout.

The same quadratic function can be written in standard, factored, or vertex form.

Standard Form to Vertex Form

This conversion needs completing the square, because that step produces the squared bracket $(x-h)^2$ . The working below shows every intermediate step.

To convert $f(x) = ax^2 + bx + c$ to vertex form $f(x) = a(x - h)^2 + k$ :

Determine the $x$ -coordinate of the vertex: $h = -\frac{b}{2a}$
Calculate the function value at the vertex: $k = f(h) = f(-\frac{b}{2a})$
Or use the formula: $k = c - \frac{b^2}{4a}$

Standard Form to Factored Form

To convert $f(x) = ax^2 + bx + c$ to factored form over the real numbers, first verify that the discriminant is nonnegative:

Determine the roots of the equation $ax^2 + bx + c = 0$ using the formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
If $D\ge0$ and the real roots are $p$ and $q$ , then:

$f(x) = a(x - p)(x - q)$

If $D<0$ , the function has no real roots and cannot be written as a product of real linear factors.

Factored Form to Standard Form

To convert $f(x) = a(x - p)(x - q)$ to standard form $f(x) = ax^2 + bx + c$ :

f(x) = a(x - p)(x - q)

By comparing with the standard form, we get:

$b = -a(p+q)$
$c = apq$

Exercises and Solutions

Each exercise gives a set of conditions and asks for the matching function. Build the function, then check that its graph passes through every given point.

First Exercise

Find the quadratic function that passes through the points $(-2, 4)$ , $(1, -5)$ , and $(3, 7)$ .

Answer:

Substitute the points into the standard equation:

\begin{aligned} 4 &= a(-2)^2 + b(-2) + c \\ &= 4a - 2b + c \end{aligned}

From the second equation, $c = -5 - a - b$ . Substitute into the first equation, then solve:

4 = 4a - 2b + (-5 - a - b)

Substitute into the third equation:

\begin{aligned} 7 &= 9(3 + b) + 3b \\ &\quad{}+ (-5 - (3 + b) - b) \end{aligned}

Then:

a = 3 + (-\frac{6}{5}) = \frac{9}{5}

The quadratic function is:

f(x) = \frac{9}{5}x^2 - \frac{6}{5}x - \frac{28}{5}

Second Exercise

Determine the quadratic function that has a vertex at $(-1, 4)$ and passes through the point $(2, -5)$ .

Answer:

Use the vertex form with $h = -1$ and $k = 4$ :

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

Substitute the point $(2, -5)$ :

-5 = a(2 + 1)^2 + 4

The quadratic function is:

\begin{aligned} f(x) &= -(x + 1)^2 + 4 \\ &= -x^2 - 2x + 3 \end{aligned}

Third Exercise

Determine the quadratic function that has roots at $x = -3$ and $x = 2$ , and has a maximum value of $4$ .

Answer:

Use the factored form with $p = -3$ and $q = 2$ :

\begin{aligned} f(x) &= a(x - (-3))(x - 2) \\ &= a(x + 3)(x - 2) \end{aligned}

Since the function has a maximum value, $a < 0$ .

The vertex is at $x = \frac{p + q}{2} = \frac{-3 + 2}{2} = -\frac{1}{2}$ .

Substitute $x = -\frac{1}{2}$ into the factored form and use the fact that the maximum value is $4$ :

\begin{aligned} f(-\frac{1}{2}) &= a(-\frac{1}{2} + 3)(-\frac{1}{2} - 2) \\ &= -a\frac{25}{4} \\ &= 4 \end{aligned}

The quadratic function is:

\begin{aligned} f(x) &= -\frac{16}{25}(x + 3)(x - 2) \\ &= -\frac{16}{25}x^2 - \frac{16}{25}x + \frac{96}{25} \end{aligned}