Polynomial Functions: Linear, Quadratic and Cubic

4. Polynomial function. $f(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ is a polynomial function of degree $n$, provided $a_0\ne0$ and every $a_i$ is real.

1. Linear function. Form: $f(x)=ax+b$ $(a\ne0)$. Example: $f(x)=-2x+3,\ x\in R$ (Fig. 6.21). Domain: $R$; Range: $R$. Properties: (1) The graph of $f(x)=ax+b$ is a straight line with slope $a$, $y$-intercept $b$, and $x$-intercept $-\dfrac{b}{a}$. (2) The function is increasing when the slope is positive and decreasing when the slope is negative.
2. Quadratic function. Form: $f(x)=ax^2+bx+c$ $(a\ne0)$ (Fig. 6.22). Domain: $R$; Range: $[k,\infty)$ for $a>0$ (where $k$ is the minimum value, at the vertex). Deriving the vertex form by completing the square: consider $y=ax^2+bx+c=a\left(x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}\right)+c-\dfrac{b^2}{4a}=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2-4ac}{4a}$, so $\left(y+\dfrac{b^2-4ac}{4a}\right)=a\left(x+\dfrac{b}{2a}\right)^2$. With the change of variable $X=x+\dfrac{b}{2a}$, $Y=y+\dfrac{b^2-4ac}{4a}$, this becomes the plain parabola $Y=aX^2$ — so the original graph is a parabola with vertex at $\left(-\dfrac{b}{2a},\ \dfrac{b^2-4ac}{4a}\right)$, equivalently $\left(-\dfrac{b}{2a},-\dfrac{D}{4a}\right)$ where $D=b^2-4ac$, opening upward (for $a>0$). Three possibilities for $a>0$, depending on the sign of the discriminant $D=b^2-4ac$: (i) If $D=0$, the parabola just touches the $X$-axis and $y\ge0$ for all $x$, e.g. $g(x)=x^2-2x+1$. (ii) If $D>0$, the parabola crosses the $X$-axis at 2 distinct points; $y$ is negative between the two roots and positive for large or small $x$. (iii) If $D<0$, the parabola lies entirely above the $X$-axis and $y\ne0$ for any $x$ — $y$ is positive for every value of $x$, e.g. $f(x)=x^2+4x+5$ (all three cases shown together in Fig. 6.23).
3. Cubic function. Example: $f(x)=ax^3+bx^2+cx+d$ $(a\ne0)$ (Fig. 6.24). Domain: $R$; Range: $R$. …