2.2 Quadratics (or Quadratic Functions)

1. The Simplest Quadratic: $y = x^2$

The graph of a quadratic is a symmetric curve known as a parabola. For the simplest baseline quadratic $y = x^2$:

$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$
$y = x^2$ $9$ $4$ $1$ $0$ $1$ $4$ $9$
V(0,0) (2, 4) (3, 9) y = x² y = -x² x y
  • The domain evaluates to $x \in \mathbb{R}$ (all real numbers).
  • The range evaluates to $y \ge 0$ (or $[0, +\infty)$).
  • The function $y = -x^2$ is an inverted reflection strictly across the x-axis, flipping the range to $y \le 0$.

2. The Quadratic Function $y = ax^2 + bx + c$

The basic characteristics of the generalized quadratic graph are as follows:

  1. Concavity: Determined by $a \ne 0$.
    • If $a > 0$, the graph is concave up (U-shaped).
    • If $a < 0$, it is concave down.
  2. Discriminant ($\Delta$): $\Delta = b^2 - 4ac$ determines the root count.
    • $\Delta > 0$: 2 distinct real roots.
    • $\Delta = 0$: 1 repeated real root.
    • $\Delta < 0$: No real roots.
  3. Roots (x-intercepts): $x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$ (exists only if $\Delta \ge 0$).
  4. y-intercept: Set $x = 0$ to obtain $y = c$.
  5. Axis of Symmetry: Given by the equation $x = -\frac{b}{2a}$. This also represents the x-coordinate of the vertex. Alternatively, if roots $x_1$ and $x_2$ are known, the vertex lies at the midpoint $x = \frac{x_1 + x_2}{2}$.

Key Features of a Parabola

y-intercept (0, c) Vertex (h, k) Root 1 Root 2 (x-intercept) (x-intercept) Axis of Symmetry x = -b / 2a O x y
a > 0 (Concave Up) Δ > 0 (2 roots) Δ = 0 (1 root) Δ < 0 (no roots) a < 0 (Concave Down) Δ > 0 (2 roots) Δ = 0 (1 root) Δ < 0 (no roots)

EXAMPLE 1

Analyze the function $y = 2x^2 - 12x + 10$.

  • $a = 2 > 0$, so the graph is concave up.
  • $\Delta = (-12)^2 - 4(2)(10) = 64 > 0$, confirming two roots.
  • Roots: $x_{1,2} = \frac{12 \pm 8}{4} \Rightarrow x = 1$ and $x = 5$.
  • y-intercept: $y = 10$.
  • Axis of symmetry: $x = -\frac{-12}{4} = 3$. At $x=3$, $y = 2(9) - 36 + 10 = -8$. The vertex is $V(3, -8)$.
V(3, -8) (0, 10) 1 5 x y

NOTICE FOR THE GDC (Casio)

  • We can find the roots $1$ and $5$ in Equation – Polynomial (degree 2).
  • We can find more characteristics in Graph mode: G-Solv (F5).
Options In our example
F1 (ROOT): for the roots $1$ and $5$
F2 (MAX) or F3 (MIN): for the vertex $(3, -8)$
F4 (YCEPT): for y-intercept $10$

3. Quadratic Inequalities

They have the form:

  • $ax^2+bx+c > 0$ or $ax^2+bx+c \ge 0$
  • $ax^2+bx+c < 0$ or $ax^2+bx+c \le 0$

If we find the roots, the graph of the function gives a clear picture of the solutions.

For example, for $2x^2 - 12x + 10 > 0$:

The roots are $1$ and $5$, the function is concave up, so it looks like:

1 5 x + + -
  • It is positive for $x < 1$ or $x > 5$. We can also write $x \in (-\infty, 1) \cup (5, +\infty)$.
  • The inequality $2x^2 - 12x + 10 \le 0$ has solutions $x \in [1, 5]$.

NOTICE:

If we are given that:

  • $ax^2+bx+c > 0$ for any $x \in \mathbb{R}$
  • $ax^2+bx+c < 0$ for any $x \in \mathbb{R}$

The graph does not intersect the x-axis, that is the quadratic has no real roots. Thus, $\Delta < 0$.

EXAMPLE 2

Let $f(x) = 2x^2 - 4x + k$. Determine the values of $k$ based on specific conditions.

  • The discriminant is $\Delta = 16 - 8k$.
  • (a) Exactly one root: $\Delta = 0 \Rightarrow 16 - 8k = 0 \Rightarrow k = 2$.
  • (b) Exactly two roots: $\Delta > 0 \Rightarrow 16 - 8k > 0 \Rightarrow k < 2$.
  • (c) No real roots: $\Delta < 0 \Rightarrow 16 - 8k < 0 \Rightarrow k > 2$.
  • (d) Has real roots (one or two): $\Delta \ge 0 \Rightarrow k \le 2$.
  • (e) $f(x) > 0$ for any $x$: Requires the parabola to float above the x-axis (no roots, concave up). $\Delta < 0 \Rightarrow k > 2$.
  • (f) $f(x) \ge 0$ for any $x$: Can touch the axis once or not at all. $\Delta \le 0 \Rightarrow k \ge 2$.

4. Forms of a Quadratic Function

  • Traditional form: $y = ax^2 + bx + c$
  • Factorization form: $y = a(x - r_1)(x - r_2)$ (where $r_1, r_2$ are roots)
  • Vertex form: $y = a(x - h)^2 + k$ (where $(h,k)$ is the vertex)

EMPHASIS ON STRATEGY: Because the vertex is universally the midpoint between the two roots, finding the roots first provides a highly efficient pathway to determining the axis of symmetry. You can quickly establish the x-coordinate of the vertex by simply averaging the roots:

$$x = \frac{r_1 + r_2}{2}$$

This property holds mathematically true universally, even when dealing with imaginary roots.

NOTICE: Finding the Vertex

  • If we know the form $y = ax^2 + bx + c$, the vertex is at $x = -\frac{b}{2a}$.
  • If we know the form $y = a(x - r_1)(x - r_2)$, that is the roots $r_1, r_2$, the vertex is at their mid-point, that is $x = \frac{r_1 + r_2}{2}$.

Since we know the x-coordinate of the vertex, that is $h$, we can also find the y-coordinate of the vertex, that is $k$. Thus we can derive the vertex form $y = a(x - h)^2 + k$.

♦ JUSTIFICATION OF THE VERTEX-FORM $y=a(x-h)^2+k$

1) The point $(h,k)$ is the vertex, i.e. a minimum or a maximum:

  • If $a>0$, then
    $a(x-h)^2 \ge 0 \quad \text{(equality holds when } x=h \text{)}$
    $\Rightarrow \quad a(x-h)^2+k \ge k$
    $\Rightarrow \quad y \ge k$
    Therefore, at $x=h$ we obtain the minimum value $y=k$.
  • If $a<0$, then
    $a(x-h)^2 \le 0 \quad \text{(equality holds when } x=h \text{)}$
    $\Rightarrow \quad a(x-h)^2+k \le k$
    $\Rightarrow \quad y \le k$
    Therefore, at $x=h$ we obtain the maximum value $y=k$.

2) Any quadratic can be expressed in the vertex form, by the "completing the square" method.

For example, for the quadratic in EXAMPLE 3 below, we can work as follows:

$$ \begin{aligned} y &= \underline{2x^2-12x}+10 \\ &= 2(x^2-6x) + 10 && \text{[only the first 2 terms]} \\ &= 2(\underline{x^2-6x+9}-9)+10 && \text{[complete the square]} \\ &= 2(x-3)^2-18+10 \\ &= 2(x-3)^2-8 \end{aligned} $$

However, it is preferable to obtain the vertex-form as in example 3 below, that is by finding the vertex $(h,k)$ and then expressing the quadratic as $y=a(x-h)^2+k$.

EXAMPLE 3

We consider again: $y = 2x^2 - 12x + 10 \quad (1)$

We find the roots: $1$ and $5$. Therefore, the factorization is:

$y = 2(x - 1)(x - 5) \quad (2)$

The vertex is at $x = -\frac{-12}{2(2)} = \frac{12}{4} = 3$
(or otherwise at the midpoint of roots: $x = \frac{1 + 5}{2} = 3$)

For $x = 3$, it is $y = -8$, hence the vertex is $(3, -8)$.

Therefore, the vertex-form of the quadratic is:

$y = 2(x - 3)^2 - 8 \quad (3)$

We may easily verify that forms (2) and (3) give (1):

  • Indeed: $y = 2(x-1)(x-5) = 2(x^2 - 5x - x + 5) = 2(x^2 - 6x + 5) = 2x^2 - 12x + 10$
  • and: $y = 2(x-3)^2 - 8 = 2(x^2 - 6x + 9) - 8 = 2x^2 - 12x + 18 - 8 = 2x^2 - 12x + 10$

EXAMPLE 4

Let: $y = -3x^2 - 15x + 42 \quad (1)$

By using the GDC, we find the roots: $-7$ and $2$. Thus the factorization is:

$y = -3(x + 7)(x - 2) \quad (2)$

We find the vertex: $V(-2.5, 60.75)$. Thus the vertex form is:

$y = -3(x + 2.5)^2 + 60.75 \quad (3)$

Notice: if you expand (2) or (3) you will obtain (1).

EXAMPLE 5

Consider $f(x) = 3x^2 + 12x$.

  • Factored form: $3x(x+4)$. The roots are immediately $x = 0$ and $x = -4$.
  • Vertex x-coordinate: utilizing the midpoint gives $x = \frac{0 + (-4)}{2} = -2$.
  • Vertex y-coordinate: $f(-2) = 3(-2)^2 + 12(-2) = 12 - 24 = -12$. The vertex is $V(-2, -12)$.
  • Vertex form: $y = 3(x+2)^2 - 12$.

EXAMPLE 6 (Polynomial with Imaginary Roots)

Let $y = 2x^2 - 8x + 14$.

We check the discriminant first: $\Delta = (-8)^2 - 4(2)(14) = 64 - 112 = -48$.
Since $\Delta < 0$, the quadratic has no real roots (the parabola floats above the x-axis). However, the property that the vertex lies at the midpoint of the roots remains universally true, even for imaginary roots!

  • Method 1: Midpoint of Imaginary Roots
    The roots evaluate to $x = \frac{8 \pm \sqrt{-48}}{4} = \frac{8 \pm 4i\sqrt{3}}{4} = 2 \pm i\sqrt{3}$.
    So, $r_1 = 2 + i\sqrt{3}$ and $r_2 = 2 - i\sqrt{3}$.
    Averaging the roots: $x = \frac{r_1 + r_2}{2} = \frac{(2 + i\sqrt{3}) + (2 - i\sqrt{3})}{2} = \frac{4}{2} = 2$.
  • Method 2: Formula ($x = -b/2a$)
    $x = -\frac{-8}{2(2)} = \frac{8}{4} = 2$.

Both methods yield the exact same x-coordinate for the vertex! Substituting $x=2$ back into the function: $y = 2(2)^2 - 8(2) + 14 = 8 - 16 + 14 = 6$.
The vertex is $V(2, 6)$, and the vertex form is $y = 2(x - 2)^2 + 6$.

5. Vieta Formulas

Given the quadratic $y = ax^2 + bx + c$ with real roots $r_1, r_2$:

  • Sum of roots (S): $S = r_1 + r_2 = -\frac{b}{a}$
  • Product of roots (P): $P = r_1 r_2 = \frac{c}{a}$

Conversely, knowing the sum and product constructs the base quadratic: $y = x^2 - Sx + P$.

EXAMPLE 7

For the function $y = 2x^2 - 12x + 10$, the roots are 1 and 5.

  • Sum: $S = -\frac{-12}{2} = 6$.
  • Product: $P = \frac{10}{2} = 5$.
  • Reconstructing: $x^2 - 6x + 5 = 0$, which scales up to the original $2x^2 - 12x + 10 = 0$.