2.2 Quadratic Functions
1. The Simplest Quadratic: $y=x^2$
The graph of a quadratic function is a symmetric curve called a parabola. The simplest quadratic function is $$y=x^2.$$ It is the parent graph for many quadratic transformations.
| $x$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|---|---|---|
| $y=x^2$ | $9$ | $4$ | $1$ | $0$ | $1$ | $4$ | $9$ |
- The domain of $y=x^2$ is $x\in\mathbb{R}$.
- The range of $y=x^2$ is $y\geq 0$, or $[0,\infty)$.
- The function $y=-x^2$ is the reflection of $y=x^2$ across the $x$-axis.
- The range of $y=-x^2$ is $y\leq 0$, or $(-\infty,0]$.
2. The Quadratic Function $y=ax^2+bx+c$
A general quadratic function has the form $$y=ax^2+bx+c,\qquad a\neq 0.$$ The coefficient $a$ controls the concavity and vertical stretch, while $b$ and $c$ affect the position of the parabola.
-
Concavity:
- If $a>0$, the graph is concave up.
- If $a<0$, the graph is concave down.
-
Discriminant:
$$\Delta=b^2-4ac.$$
The discriminant determines how many real roots the quadratic has.
- If $\Delta>0$, there are two distinct real roots.
- If $\Delta=0$, there is one repeated real root.
- If $\Delta<0$, there are no real roots.
- Roots / $x$-intercepts: $$x=\dfrac{-b\pm\sqrt{\Delta}}{2a},\qquad \Delta\geq 0.$$
- $y$-intercept: set $x=0$. Since $$y=a(0)^2+b(0)+c=c,$$ the $y$-intercept is $(0,c)$.
- Axis of symmetry: $$x=-\dfrac{b}{2a}.$$ This is also the $x$-coordinate of the vertex.
- Vertex from roots: if the roots are $r_1$ and $r_2$, then the axis of symmetry is $$x=\dfrac{r_1+r_2}{2}.$$
Key Features of a Parabola
Discriminant Cases
EXAMPLE 1
Analyze the quadratic function $$y=2x^2-12x+10.$$
NOTICE FOR THE GDC (Casio)
- Use Equation $\rightarrow$ Polynomial $\rightarrow$ degree 2 to find the roots.
- Use Graph mode $\rightarrow$ G-Solv (F5) to find graph features.
| Option | Purpose | In this example |
|---|---|---|
| F1 ROOT | Find roots / $x$-intercepts | $1$ and $5$ |
| F2 MAX / F3 MIN | Find the vertex | $(3,-8)$ |
| F4 Y-CEPT | Find the $y$-intercept | $10$ |
3. Quadratic Inequalities
Quadratic inequalities have one of the following forms:
- $ax^2+bx+c>0$ or $ax^2+bx+c\geq 0$
- $ax^2+bx+c<0$ or $ax^2+bx+c\leq 0$
Once the roots are known, the graph shows where the quadratic is positive, negative, or zero.
For $$2x^2-12x+10>0,$$ the roots are $1$ and $5$, and the parabola is concave up.
- The expression is positive for $$x<1\qquad\text{or}\qquad x>5.$$ Hence, $$x\in(-\infty,1)\cup(5,\infty).$$
- The inequality $$2x^2-12x+10\leq 0$$ has solution $$x\in[1,5].$$
NOTICE
If $$ax^2+bx+c>0\quad\text{for every }x\in\mathbb{R},$$ or $$ax^2+bx+c<0\quad\text{for every }x\in\mathbb{R},$$ then the graph does not intersect the $x$-axis. Therefore, the quadratic has no real roots, so $$\Delta<0.$$
EXAMPLE 2
Let $$f(x)=2x^2-4x+k.$$ Determine the values of $k$ for each condition.
- Exactly one root: $$\Delta=0\quad\Longrightarrow\quad 16-8k=0\quad\Longrightarrow\quad k=2.$$
- Exactly two roots: $$\Delta>0\quad\Longrightarrow\quad 16-8k>0\quad\Longrightarrow\quad k<2.$$
- No real roots: $$\Delta<0\quad\Longrightarrow\quad 16-8k<0\quad\Longrightarrow\quad k>2.$$
- Has real roots: $$\Delta\geq 0\quad\Longrightarrow\quad k\leq 2.$$
- $f(x)>0$ for every $x\in\mathbb{R}$: Since $a=2>0$, the parabola opens upward. To stay strictly above the $x$-axis, it must have no real roots: $$\Delta<0\quad\Longrightarrow\quad k>2.$$
- $f(x)\geq 0$ for every $x\in\mathbb{R}$: The parabola may touch the $x$-axis once or float above it: $$\Delta\leq 0\quad\Longrightarrow\quad k\geq 2.$$
4. Forms of a Quadratic Function
- Traditional form: $$y=ax^2+bx+c.$$
- Factorized form: $$y=a(x-r_1)(x-r_2),$$ where $r_1$ and $r_2$ are the roots.
- Vertex form: $$y=a(x-h)^2+k,$$ where $(h,k)$ is the vertex.
EMPHASIS ON STRATEGY: If the roots are known, the $x$-coordinate of the vertex is the midpoint of the roots:
$$x=\dfrac{r_1+r_2}{2}.$$
This midpoint property also agrees with the formula $$x=-\dfrac{b}{2a}.$$
NOTICE: Finding the Vertex
- From $y=ax^2+bx+c$, use $$x=-\dfrac{b}{2a}.$$
- From $y=a(x-r_1)(x-r_2)$, use $$x=\dfrac{r_1+r_2}{2}.$$
- After finding the $x$-coordinate $h$, substitute $x=h$ into the function to find $k$.
- Then write the quadratic as $$y=a(x-h)^2+k.$$
♦ JUSTIFICATION OF THE VERTEX FORM $y=a(x-h)^2+k$
1) The point $(h,k)$ is the vertex.
- If $a>0$, then $$a(x-h)^2\geq 0.$$ Therefore, $$a(x-h)^2+k\geq k.$$ Hence $y\geq k$, and the minimum value is $k$ at $x=h$.
- If $a<0$, then $$a(x-h)^2\leq 0.$$ Therefore, $$a(x-h)^2+k\leq k.$$ Hence $y\leq k$, and the maximum value is $k$ at $x=h$.
2) Any quadratic can be written in vertex form by completing the square. For example:
This agrees with the vertex $(3,-8)$ found earlier.
EXAMPLE 3
Express $$y=2x^2-12x+10$$ in factorized form and vertex form.
EXAMPLE 4
Let $$y=-3x^2-15x+42.$$ Express it in factorized form and vertex form.
EXAMPLE 5
Consider $$f(x)=3x^2+12x.$$ Find the roots, vertex, and vertex form.
EXAMPLE 6 (Polynomial with Imaginary Roots)
Let $$y=2x^2-8x+14.$$ Find the vertex and vertex form.
5. Vieta Formulas
Given the quadratic $$y=ax^2+bx+c,$$ with roots $r_1$ and $r_2$, Vieta's formulas state:
- Sum of roots: $$S=r_1+r_2=-\dfrac{b}{a}.$$
- Product of roots: $$P=r_1r_2=\dfrac{c}{a}.$$
Conversely, if a monic quadratic has sum of roots $S$ and product of roots $P$, then $$x^2-Sx+P=0.$$
EXAMPLE 7
For $$y=2x^2-12x+10,$$ the roots are $1$ and $5$. Verify Vieta's formulas.