$\blacktriangleright$ $S(x) = A \, x^2 + B \, x + C$ 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.

Vertex form comes from completing the square. It gets its name from the graph of quadratic functions.

Graphs of Quadratic Functions

Quadratic functions all have parabolas for graphs.

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.

$$V(x) = A \, (x - h)^2 + k$$

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.

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

$f$ is a quadratic function and will have a parabola for a graph.

Analysis

We are given $f(x)$ in standard form. Let’s get the other two forms.

Factored form: $f(x) = (x+5)(x-1)$.

Vertex form: $f(x) = (x+2)^2 - 9$

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 as the intercepts $(-5, 0)$ and $(1,0)$.

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

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

The range of $f$ is $[-9,\infty )$.

There are no discontinuities or singularities.