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Learn to construct quadratic functions using three points, vertex form, roots, and symmetry. Learn different methods with worked examples and solutions.

---

## How to Determine Quadratic Function Equations

Constructing a quadratic function means determining the form of the equation $$f(x) = ax^2 + bx + c$$ based on given information. There are several different ways to construct a quadratic function, depending on the available information.

Visible text: Constructing a quadratic function means determining the form of the equation based on given information. There are several different ways to construct a quadratic function, depending on the available information.

## Forms of Quadratic Functions

Before we start constructing quadratic functions, let's understand the three forms of quadratic functions:

1. **Standard Form**: $$f(x) = ax^2 + bx + c$$
2. **Factored Form**: $$f(x) = a(x - p)(x - q)$$ where $$p$$ and $$q$$ are the roots of the equation
3. **Vertex Form**: $$f(x) = a(x - h)^2 + k$$ where $$(h, k)$$ is the vertex point

Visible text: 1. **Standard Form**: 
2. **Factored Form**: where and are the roots of the equation
3. **Vertex Form**: where is the vertex point

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

## Methods of Constructing Quadratic Functions

### Three Points

The most common way to construct a quadratic function is by using three known points on the curve.

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 $$c$$.

Visible text: If we have three points , , and , we can substitute these values into the standard equation to get three linear equations with three variables , , and .

Component: MathContainer
Children:

```math
y_1 = ax_1^2 + bx_1 + c
```

```math
y_2 = ax_2^2 + bx_2 + c
```

```math
y_3 = ax_3^2 + bx_3 + c
```

**Example**:

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

Visible text: Find the quadratic function that passes through points , , and .

Substitute the coordinate values into the standard equation.

Component: MathContainer
Children:

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

```math
L(0, -3): -3 = a(0)^2 + b(0) + c
```

```math
M(1, -4): -4 = a(1)^2 + b(1) + c
```

Simplify these equations.

Component: MathContainer
Children:

```math
0 = a - b + c
```

```math
-3 = c
```

```math
-4 = a + b + c
```

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

Visible text: Substitute into the first and third equations.

Component: MathContainer
Children:

```math
0 = a - b - 3
```

```math
-4 = a + b - 3
```

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

Visible text: Simplify and add the two equations to find the value of .

Component: MathContainer
Children:

```math
a - b = 3
```

```math
a + b = -1
```

```math
2a = 2
```

```math
a = 1
```

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

Visible text: Substitute the value of into the equation to find the value of .

Component: MathContainer
Children:

```math
1 - b = 3
```

```math
b = -2
```

The resulting quadratic function is:

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

Component: ContentBlock
Children:
Component: LineEquation
Props:
- title: Graph of $$f(x) = x^2 - 2x - 3$$
  Visible text: Graph of
- description: Parabola passing through points $$K(-1, 0)$$,{" "}
$$L(0, -3)$$, and $$M(1, -4)$$.
  Visible text: Parabola passing through points ,{" "}
, and .
- cameraPosition: [2, 3, 8]
- data: [
{
points: Array.from({ length: 7 }, (_, i) => {
const x = i - 3; // x values from -3 to 3
return { x, y: x * x - 2 * x - 3, z: 0 };
}),
color: getColor("TEAL"),
labels: [
{
text: "f(x) = x² - 2x - 3",
at: 5,
offset: [2, 0.3, 0],
},
{
text: "L(0, -3)",
at: 3,
offset: [-1, 0, 0],
},
{
text: "M(1, -4)",
at: 4,
offset: [0, -0.5, 0],
},
{
text: "K(-1, 0)",
at: 2,
offset: [-1, 0.3, 0],
},
],
},
]

### 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 $$a$$.

Visible text: If we know the coordinates of the vertex and another point on the graph, we can use the vertex form to find the value of .

**Example**:

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

Visible text: Find the quadratic function that has a vertex at and passes through the point .

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

Visible text: Use the vertex form with and .

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

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

Visible text: Substitute the point to find the value of .

Component: MathContainer
Children:

```math
4 = a(4 - 2)^2
```

```math
4 = 4a
```

```math
a = 1
```

The resulting quadratic function is:

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

Component: ContentBlock
Children:
Component: LineEquation
Props:
- title: Graph of $$f(x) = (x - 2)^2$$
  Visible text: Graph of
- description: Parabola with vertex at $$(2, 0)$$ and passing through
point $$(4, 4)$$.
  Visible text: Parabola with vertex at and passing through
point .
- cameraPosition: [6, 6, 8]
- data: [
{
points: Array.from({ length: 9 }, (_, i) => {
const x = i - 1; // x values from -1 to 7
return { x, y: Math.pow(x - 2, 2), z: 0 };
}),
color: getColor("INDIGO"),
labels: [
{
text: "f(x) = (x - 2)²",
at: 7,
offset: [0.3, 0.3, 0],
},
{
text: "(2, 0)",
at: 3,
offset: [0, 0.3, 0],
},
{
text: "(4, 4)",
at: 5,
offset: [0.3, 0.3, 0],
},
],
},
]

### Equation Roots

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

Visible text: If we know the roots (-intercepts) and of the quadratic function, and an additional point on the curve, we can use the factored form .

**Example**:

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

Visible text: Find the quadratic function that has roots at and , and passes through the point .

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

Visible text: Use the factored form with and .

```math
f(x) = a(x - (-2))(x - 3) = a(x + 2)(x - 3)
```

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

Visible text: Substitute the point to find the value of .

Component: MathContainer
Children:

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

```math
-6 = a(3)(-2)
```

```math
-6 = -6a
```

```math
a = 1
```

The resulting quadratic function is:

```math
f(x) = (x + 2)(x - 3) = x^2 - x - 6
```

Component: ContentBlock
Children:
Component: LineEquation
Props:
- title: Graph of $$f(x) = (x + 2)(x - 3)$$
  Visible text: Graph of
- description: Parabola with roots at $$x = -2$$ and{" "}
$$x = 3$$, passing through point{" "}
$$(1, -6)$$.
  Visible text: Parabola with roots at and{" "}
, passing through point{" "}
.
- cameraPosition: [4, 5, 10]
- data: [
{
points: Array.from({ length: 11 }, (_, i) => {
const x = i - 4; // x values from -4 to 6
return { x, y: (x + 2) * (x - 3), z: 0 };
}),
color: getColor("ORANGE"),
labels: [
{
text: "f(x) = (x + 2)(x - 3)",
at: 6,
offset: [3, 0.3, 0],
},
{
text: "(-2, 0)",
at: 2,
offset: [0.3, 0.3, 0],
},
{
text: "(3, 0)",
at: 7,
offset: [0.3, 0.3, 0],
},
{
text: "(1, -6)",
at: 5,
offset: [0.3, -0.3, 0],
},
],
},
]

### Axis of Symmetry and Discriminant

We can also construct a quadratic function by knowing the axis of symmetry (or the $$x$$-coordinate of the vertex) and the value of the discriminant.

Visible text: We can also construct a quadratic function by knowing the axis of symmetry (or the -coordinate of the vertex) and the value of the discriminant.

**Example**:

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

Visible text: Find a quadratic function with axis of symmetry and discriminant .

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

Visible text: From the axis of symmetry , we know that , so .

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

Visible text: From the discriminant , we know that .

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

Visible text: Substitute into the discriminant equation.

Component: MathContainer
Children:

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

```math
4a^2 - 4ac = 16
```

```math
4a(a - c) = 16
```

```math
a(a - c) = 4
```

There are many values of $$a$$ and $$c$$ that satisfy this equation. Let's take the simple case with $$a = 1$$.

Visible text: There are many values of and that satisfy this equation. Let's take the simple case with .

Component: MathContainer
Children:

```math
1(1 - c) = 4
```

```math
1 - c = 4
```

```math
c = -3
```

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

Visible text: With , , and , the resulting quadratic function is:

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

### Symmetric Coordinates

We can construct a quadratic function by utilizing the symmetry property of parabolas.

**Example**:

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

Visible text: Determine the quadratic function that passes through the points , , and .

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 $$(0, 0)$$.

Visible text: Since the points and have the same -value and are at the same distance from the -axis, the curve is symmetric about the -axis. This means the axis of symmetry is , and the vertex is at .

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

Visible text: Use the vertex form with and .

```math
f(x) = ax^2
```

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

Visible text: Substitute the point to find the value of .

Component: MathContainer
Children:

```math
1 = a(4)^2
```

```math
1 = 16a
```

```math
a = \frac{1}{16}
```

The resulting quadratic function is:

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

Component: ContentBlock
Children:
Component: LineEquation
Props:
- title: Graph of $$f(x) = \frac{1}{16}x^2$$
  Visible text: Graph of
- description: Parabola passing through points $$(0, 0)$$,{" "}
$$(4, 1)$$, and $$(-4, 1)$$.
  Visible text: Parabola passing through points ,{" "}
, and .
- data: [
{
points: Array.from({ length: 13 }, (_, i) => {
const x = i - 6; // x values from -6 to 6
return { x, y: (1 / 16) * x * x, z: 0 };
}),
color: getColor("TEAL"),
labels: [
{
text: "f(x) = (1/16)x²",
at: 11,
offset: [-2, 0.3, 0],
},
{
text: "(0, 0)",
at: 6,
offset: [0.3, 0.3, 0],
},
{
text: "(4, 1)",
at: 10,
offset: [0.3, 0.3, 0],
},
{
text: "(-4, 1)",
at: 2,
offset: [0.3, 0.3, 0],
},
],
},
]

## Transformations Between Quadratic Function Forms

### Standard Form to Vertex Form

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

Visible text: To convert to vertex form :

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

Visible text: 1. Determine the -coordinate of the vertex: 
2. Calculate the function value at the vertex: 
3. Or use the formula:

### Standard Form to Factored Form

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

Visible text: To convert to factored form :

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

   
   
   ```math
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   ```

2. If the roots are $$p$$ and $$q$$, then:

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

Visible text: 1. Determine the roots of the equation using the formula:

 
 

2. If the roots are and , then:

### Factored Form to Standard Form

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

Visible text: To convert to standard form :

Component: MathContainer
Children:

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

```math
f(x) = a(x^2 - qx - px + pq)
```

```math
f(x) = a(x^2 - (p+q)x + pq)
```

```math
f(x) = ax^2 - a(p+q)x + apq
```

By comparing with the standard form, we get:

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

Visible text: - 
-

## Exercises and Solutions

### First Exercise

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

Visible text: Find the quadratic function that passes through the points , , and .

**Answer**:

Substitute the points into the standard equation:

Component: MathContainer
Children:

```math
4 = a(-2)^2 + b(-2) + c = 4a - 2b + c
```

```math
-5 = a(1)^2 + b(1) + c = a + b + c
```

```math
7 = a(3)^2 + b(3) + c = 9a + 3b + c
```

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

Visible text: From the second equation, .
Substitute into the first equation, then solve:

Component: MathContainer
Children:

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

```math
9 = 3a - 3b
```

```math
3 = a - b
```

```math
a = 3 + b
```

Substitute into the third equation:

Component: MathContainer
Children:

```math
7 = 9(3 + b) + 3b + (-5 - (3 + b) - b)
```

```math
7 = 19 + 10b
```

```math
b = -\frac{12}{10} = -\frac{6}{5}
```

Then:

Component: MathContainer
Children:

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

```math
c = -5 - \frac{9}{5} - (-\frac{6}{5}) = -\frac{28}{5}
```

Therefore, the quadratic function is:

```math
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 $$(2, -5)$$.

Visible text: Determine the quadratic function that has a vertex at and passes through the point .

**Answer**:

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

Visible text: Use the vertex form with and :

```math
f(x) = a(x - (-1))^2 + 4 = a(x + 1)^2 + 4
```

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

Visible text: Substitute the point :

Component: MathContainer
Children:

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

```math
-5 = 9a + 4
```

```math
-9 = 9a
```

```math
a = -1
```

Therefore, the quadratic function is:

```math
f(x) = -(x + 1)^2 + 4 = -x^2 - 2x + 3
```

### Third Exercise

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

Visible text: Determine the quadratic function that has roots at and , and has a maximum value of .

**Answer**:

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

Visible text: Use the factored form with and :

```math
f(x) = a(x - (-3))(x - 2) = a(x + 3)(x - 2)
```

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

Visible text: Since the function has a maximum value, .

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

Visible text: The vertex is at .

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

Visible text: Substitute into the factored form and use the fact that the maximum value is :

Component: MathContainer
Children:

```math
f(-\frac{1}{2}) = a(-\frac{1}{2} + 3)(-\frac{1}{2} - 2) = -a\frac{25}{4} = 4
```

```math
a = -\frac{16}{25}
```

Therefore, the quadratic function is:

```math
f(x) = -\frac{16}{25}(x + 3)(x - 2) = -\frac{16}{25}x^2 - \frac{16}{25}x + \frac{96}{25}
```