We have three forms for quadratic functions (or equations):

\(\blacktriangleright \) \(S(x) = A \, x^2 + B \, x + C\) with \(A \ne 0\) : Standard Form.

\(\blacktriangleright \) \(F(t) = A \, (t - r_1)(t - r_2)\) with \(A \ne 0\) : Factored Form.

\(\blacktriangleright \) \(V(d) = A \, (d - h)^2 + k\) with \(A \ne 0\) : Vertex Form.

The standard form is the default form.

Factored form is the most helpful, because much analysis comes from a function’s zeros.

Vertex form gets its name from the graph of a quadratic function, which is a parabola and has a vertex.

We want both graphical and algebraic thinking. A graph is inherrantly inaccurate and so we want our analysis reasoning to have algebraic and functional explanations. But, we still want to use graphs to help our algebra.

Graphs of Quadratic Functions

Quadratic functions all have parabolas for graphs.

Graph Description: Parabola opening down. Top point (vertex) at \((1,6.5)\). Intercepts are \((-4,0)\) and \((6,0)\).

Graph Description: Parabola opening up. Bottom point (vertex) at \((-1,-6.5)\). Intercepts are \((-6,0)\) and \((4,0)\).

The extreme point on a parabola is called the vertex. It is the lowest or highest point on the parabola, depending on whether the parabola opens up or down.

This can be seen from the vertex form of the formula.

The squared term, \(A \, (x - h)^2\) has the same sign as \(A\), except when it equals \(0\). That happens at \(h\). When \(x = h\), then \(V(h) = k\), which is either the minimum or maximum value of \(V\). The vertex is the graphical representation of the the extreme value of the quadratic function and where this extrema occurs in the domain.

In addition, the intercepts represent the zeros of the quadratic function and we have seen there can be \(0\), \(1\), or \(2\) real zeros for a quadratic function. Therefore, there can be \(0\), \(1\), or \(2\) intercepts for a parabola.

Analysis

The first step in analyzing a quadratic function is categorizing it as a quadratic. To do this, you need to show that the given formula is equivalent to our official template for quadratic functions.

Show the given function CAN be rewritten in the form of the official template.

Once you have categorized the function as quadratic then

• the domain of the function is all real numbers
• the function has exactly \(0\), \(1\), or \(2\) zeros
• the function is continuous with no singularities
• the end-behavior on both sides is the same and it is either \(-\infty \) or \(\infty \) depending on the sign of the leading coefficient
• the function either increases and then decreases or decreases and then increases depending on the sign of the leading coeffcient
• the function either has a global maximum or minimum depending on the sign of the leading coeffcient
• the function has one local extrema and it is the same as the global extrema
• the range looks like either \((-\infty , m]\) or \([m, \infty )\) depending on the sign of the leading coeffcient

That is a lot of free analysis information gained from categorizing the function.

Analysis

Analyze \(f(x) = x^2 + 4 \, x - 5\) with its natural domain.

explanation

Category

\(f\) is a quadratic function, since its formula matches the standard form, \(A \, x^2 + B \, x + C\).

Domain

\(f\) is a quadratic function, therefore its domain is \((-\infty , \infty )\).

Zeros

We can use the quadratic formula to get the zeros.

\[ \frac {-4 \pm \sqrt {4^2 - 4(1)(-5)}}{2(1)} = \frac {-4 \pm \sqrt {36}}{2} = \frac {-4 \pm 6}{2} \]

This give \(-5\) and \(1\) as the zeros.

This gives us the factored for : \(f(x) = (x+5)(x-1)\).

It might have been quick to use the distributive property to factor the quadratic instead of using the quadratic formula.

From the factored form, we can also see that \(-5\) and \(1\) are the zeros of \(f(x)\).

Continuity

\(f\) is a quadratic function, therefore it is continuous. Quadratic functions do not have singularities.

End-Behavior

\(f\) is a quadratic function, with a positive leading coefficient. Therefore, its end-behavior is unbounded positively in both directions.

\[ \lim \limits _{x \to -\infty } f(x) = \infty \]

\[ \lim \limits _{x \to \infty } f(x) = \infty \]

Behavior (increasing and decreasing)

\(f\) is a quadratic function, with a positive leading coefficient. Therefore, it will decrease and then increase. It switches behavior at the domain number for the vertex, which is

\[ -\frac {b}{2a} = -\frac {4}{2(1)} = -2 \]

(\(-2\) is called the critical number of \(f\).)

\(f\) decreases on \((-\infty , -2)\) and increases on \((-2, \infty )\).

Note: We are just beginning, so parentheses are ok. Really, \(f\) decreases on \((-\infty , -2]\) and increases on \([-2, \infty )\).

Global and Local Maximum and Minimum

For a quadratic function, the global and local extreme values are the same.

Since \(f\) decreases on \((-\infty , -2)\) and increases on \((-2, \infty )\), we have a global minimum at \(-2\). The minimum value is \(f(-2) = -9\). This is also a local minimum and is the only local extrema.

\(\lim \limits _{x \to -\infty } f(x) = \infty \) tells us that there is no global maximum.

Range

\(f\) is continuous, with a global minimum and unbounded positively.

The range is \([-9, \infty )\).

A Nice Graph

\(f\) is a quadratic function and will have a parabola for a graph.

From any of the three forms, we can see that the leading coefficient is \(1\), which is positive. Thus, our parabola will open up.

From the factored form, we can see that \(-5\) and \(1\) are the zeros of \(f(x)\). These will be represented by the intercepts \((-5, 0)\) and \((1,0)\).

From the vertex form, we can see that the lowest point of the parabola will be \((-2, 9)\). Or, the function \(f\) has a global minimum value of \(-9\), which occurs at \(-2\) in the domain.

\(\blacktriangleright \) desmos graph