## Summary

IDefinition, domain and range, graphical representationIIForms of a quadratic functionAStandard form of a quadratic functionBVertex form of a quadratic functionCIntercept form of a quadratic functionIIIAverage rate of change and difference quotientIVQuadratic equations and inequalities## Definition, domain and range, graphical representation

### Quadratic function

A quadratic function is any function of the following form:

**f(x)=ax^2+bx+c**

a,b,c are all real numbers and a\not =0.

The following functions are quadratic:

- f(x)=x^2+2x+3
- g(x)=x^2
- h(x)=4x^2-2x

The graph of a quadratic function is a parabola.

If f(x)=ax^2+bx+c, then:

- The parabola faces up if a>0.
- The parabola faces down if a<0.

### Vertex

The vertex of a parabola is the point where the parabola switches from increasing to decreasing (or from decreasing to increasing).

If f(x)=ax^2+bx+c is a quadratic function then the range of f(x) is determined by the vertex of the parabola.

## Forms of a quadratic function

A quadratic function can be presented in three forms:

- standard form
- vertex form
- intercept form

### Standard form of a quadratic function

#### Standard form of a quadratic function

The standard form of a quadratic function is:

**f(x)=ax^2+bx+c**

The following quadratic equation is in standard form:

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

### Vertex form of a quadratic function

#### Vertex form of a quadratic function

The vertex form of a quadratic function is:

f\left(x\right)=a\left(x-h\right)^2+k

a,h,k are real numbers.

Observe the following:

f\left(x\right)=a\left(x-h\right)^2+k=a\left(x^2-2hx+h^2\right)+k=ax^2-\left(2ah\right)x+\left(ah^2+c\right)

It is indeed a quadratic function.

The vertex form of a quadratic function is very useful.

Consider the following function:

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

If it is in vertex form then the point (h,k) is the vertex of the parabola. It follows that the range of f(x) is either:

- [k,\infty) if a>0
- (-\infty,k] if a<0

Consider the following quadratic function:

f(x)=-3(x-2)^2+7

The vertex of the quadratic function is at (2,-7) and the range of f(x) is (-\infty, 7].

Consider the following function:

f(x)=2(x-4)^2+17

The quadratic function has a vertex at (4{,}17) and the range of f(x) is [17,\infty).

Suppose f(x)=ax^2+bx+c is a quadratic function in standard form. Then the vertex form of f(x) is:

**f(x)=a\left(x+\dfrac{b}{2a}\right)^2+c-\dfrac{b^2}{4a}**

In particular, the vertex of f(x) is at \left(\dfrac{-b}{2a}, c-\dfrac{b^2}{4a}\right)

- If a>0 then the range of f(x) is \left[c-\dfrac{b^2}{4a},\infty\right)
- If a<0 then the range of f(x) is \left(-\infty,c-\dfrac{b^2}{4a}\right]

Consider the following function:

Let f(x)=x^2-4x+3

Then f(x) is in standard form with :

- a=1
- b=-4
- c=3

The vertex form of f(x) is then computed as:

f\left(x\right)=a\left(x+\dfrac{b}{2a}\right)^2+c-\dfrac{b^2}{4a}\\=\left(x+\dfrac{-4}{2}\right)^2+3-\dfrac{16}{4}\\= \left(x-2\right)^2+3-4\\=\left(x-2\right)^2-1

Therefore the vertex of f(x) is at (2,-1) and the range of f(x) is [-1,\infty).

### Intercept form of a quadratic function

#### x-Intercept form

The x-intercept form of a quadratic equation is:

** f\left(x\right)=a\left(x-\alpha_1\right)\left(x-\alpha_2\right) **

Where a,\alpha_1, \alpha_2 are real numbers.

The following quadratic function is in intercept form:

f(x)=(x-1)(x+1)

The following quadratic function is in intercept form:

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

The following quadratic function is in intercept form:

f(x)=-7(x-1)x

If f\left(x\right)=a\left(x-\alpha_1\right)\left(x-\alpha_2\right) is a quadratic function in x -intercept form, then the points \left(\alpha_1{,}0\right) and \left(\alpha_2{,}0\right) are the x -intercepts of the graph of f\left(x\right).

The quadratic function f(x)=2(x-1)(x+17) has x-intercepts at (1{,}0) and (-17{,}0).

#### Discriminant

Let f\left(x\right)=ax^2+bx+c be a quadratic function in standard form. The discriminant of f\left(x\right) is:

** b^2-4ac **

Consider the following quadratic function:

f\left(x\right)=2x^2-7x+1

The discriminant of f\left(x\right) is:

\left(-7\right)^2-4\left(2\right)\left(1\right)=49-8=41

Consider the following quadratic function:

f(x)=ax^2+bc+c

Let d=b^2-4ac be the discriminant of f\left(x\right). We then have the following:

- If d \gt 0 then f\left(x\right) has two x -intercepts.
- If d=0 then f\left(x\right) has one x -intercept.
- If d \lt 0 then f\left(x\right) does not have an x -intercept.

Furthermore, if d\geq 0 then the x -intercept form of f\left(x\right) is:

f(x)=ax^2+bx+c=a(x-\alpha_1) (x-\alpha_2)

Where:

- \alpha_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}
- \alpha_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}

The above formulas for finding the x-values of the x-intercept form of a quadratic equation is known as the **quadratic formula.**

Consider the following quadratic function:

f(x)=x^2-3x-10

To find its x-intercept form we factor the quadratic x^2-3x-10 as follows:

x^2-3x-10=(x-\alpha_1)(x-\alpha_2)

We have:

- \alpha_1=\dfrac{-(-3)+\sqrt{(-3)^2-4(1)(-10)}}{2(1)}=\dfrac{3+\sqrt{9+40}}{2}=\dfrac{3+7}{2}=5
- \alpha_2=\dfrac{-(-3)-\sqrt{(-3)^2-4(1)(-10)}}{2(1)}=\dfrac{3-7}{2}=-2

Therefore the quadratic function has x-intercept form that is:

f(x)=1(x-5)(x-(-2))=(x-5)(x+2)

Consider the following quadratic function:

f(x)=x^2+1

It does not have an x-intercept form. This can be seen in one of two ways.

First, the square of any real number is nonnegative and therefore x^2+1\geq 1 for all x, hence the graph of f(x) does not have an x-intercept.

Also, compute the discriminant:

d=0^2-4\left(1\right)\left(1\right)=-4

d\lt0, therefore the function has no x-intercept.

Suppose f\left(x\right)=x^2+bx+c has x -intercept form f\left(x\right)=\left(x-\alpha_1\right)\left(x-\alpha_2\right). Observe that:

\left(x-\alpha_1\right)\left(x-\alpha_2\right)=x^2-\left(\alpha_1+\alpha_2\right)x+\alpha_1\alpha_2

Therefore we have:

- \alpha_1+\alpha_2=-b
- \alpha_1\alpha_2=c

Thus when factoring a quadratic function, it is not necessary to use the quadratic formula. Instead we can try to find \alpha_1 and \alpha_2 satisfying the above equations.

Let f(x) be the following quadratic equation:

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

To factor this equation notice:

- (-2)\cdot (-2)=4
- (-2)+(-2)=-4

Therefore the quadratic equation factors as follows:

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

The graph of f(x) has one x-intercept at (2{,}0).

Suppose f(x)=x^2-c is a quadratic equation. Then f(x)=x^2+0b-c. Observe the following:

- \sqrt{c}+(-\sqrt{c})=0
- \sqrt{c}\cdot (-\sqrt{c})=-c

Therefore the quadratic equation has the following x-intercept form:

**f(x)=x^2-c=\left(x-\sqrt{c}\right)\left(x+\sqrt{c}\right)**

The quadratic equation f(x)=x^2-49 has x-intercept form:

f(x)=x^2-49=(x-7)(x+7)

## Average rate of change and difference quotient

### Average rate of change of a quadratic function

Let f(x)=ax^2+bx+c be a quadratic function. The average rate of change of f(x) from two real numbers x_0 to x_1 is :

**\dfrac{f(x_1)-f(x_0)}{x_1-x_0}=a(x_0+x_1)+b**

Let f(x)=x^2+7x+3. The average rate of change of f(x) from 3 to 5 is:

\dfrac{f\left(5\right)-f\left(3\right)}{5-3}\\=1\left(5+3\right)+7\\=8+7\\=15

### Difference quotient of a quadratic function

Let f(x)=ax^2+bx+c be a quadratic function. The difference quotient of f(x) is:

**\dfrac{f(x+h)-f(x)}{h}=2ax+b +ah**

Consider the following function:

f(x)=x^2+7x+3

The difference quotient of f(x) is:

\dfrac{f\left(x+h\right)-f\left(x\right)}{h}\\=2\left(1\right)x+7+\left(1\right)h\\=2x+7+h

## Quadratic equations and inequalities

### Quadratic function

A quadratic equation is any equation equivalent to an equation of the following form:

** ax^2+bx+c=0 **

The following is a quadratic equation:

2x^2+7x-3=0

Consider the following quadratic equation:

a(x-b)(x-c)=0

The solutions are:

- x=b
- x=c

Consider the quadratic equation:

(x-1)(x+3)=0

The solutions are:

- x=1
- x=-3

Consider the following equation:

x^2+7=3x+5

To solve it we begin by moving all terms to one side of the equation:

x^2+7-3x-5=0

x^2-3x+2=0

We then find the x-intercept form of the quadratic:

x^2-3x+2

By observing :

- (-2)\cdot (-1)=2
- (-2)+(-1)=-3

The quadratic equation is then solved as follows:

- x^2-3x+2=0
- (x-2)(x-1)=0

So the solutions to the equation are x=2 and x=1.

Not all quadratic equations have a solution.

Consider the following quadratic equation:

x^2+4=0

There is no solution because it does not have an x-intercept form.

### Quadratic Inequality

A quadratic inequality is any inequality equivalent to an inequality of the following forms:

**ax^2+bx+c\geq 0****ax^2+bx+c\gt 0****ax^2+bx+c\leq 0****ax^2+bx+c\lt 0**

The following inequality is a quadratic inequality.

2x^2+3x-1\lt 0

Quadratic inequalities are solved by finding the x-intercept form of the quadratic.

Consider the following quadratic function:

f(x)=a(x-b)(x-c)

Suppose a>0 and b\leq c. The graph of the quadratic function has x-intercepts at (b,0) and (c,0).

- f(x) is only negative when b<x<c.
- f(x) is positive when x<b and when x>c.

Consider the following quadratic inequality:

2(x-1)(x+3)\leq 0

- When x=1 or x=-3, the quadratic is 0
- When x is between -3 and 1, the quadratic is less than 0

Therefore it has the following solution set:

-3\leq x \leq 1

Consider the following inequality:

x^2+7\gt 3x+5

It is solved as follows:

x^2+7\gt 3x+5

x^2+7-3x-5\gt 0

x^2-3x+2\gt 0

(x-1)(x-2)\gt 0

The solution set to the inequality is therefore:

(-\infty,1)\cup (2,\infty)

Consider the following quadratic inequality:

x^2+1\geq 0

The solution set to the quadratic inequality is all real numbers since the graph of f(x)=x^2+1 lies above the x-axis.

Consider the quadratic inequality:

x^2+1\lt 0

It has no solution because the function f(x)=x^2+1 is always positive.