202110120936 Exercises 13.3.6 (Q43)

Graph each function by algebraically determining its key features. Then state the domain and range of the function.

$f(x)=x^2-7x+12$

Solution.

$\begin{aligned} f(x) & = x^2-7x+12 \\ & = (x-3)(x-4) \end{aligned}$

We see that $y=f(x)=0$ when $x=\{ 3,4\}$ .

i. $\therefore$ The $x$ -intercepts of $f(x)$ are thus $(3,0)$ and $(4,0)$ .

Plugging in $x=0$ will give the $y$ -intercept:

$\begin{aligned} y\textrm{-intercept}& = f(0) \\ & = (0)^2-7(0)+12 \\ & = 12 \end{aligned}$

ii. $\therefore$ The $y$ -intercept is thus $(0,12)$ .

Differentiating $y=f(x)$ with respect to $x$ ,

$\begin{aligned} & \quad f'(x) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^2-7x+12) \\ & = 2x-7 \\ \end{aligned}$

When $x=3.5$ and $f(3.5)=-0.25$ , the slope of $f(x)$ is zero, i.e., $f'(x)=0$ .

iii. $\therefore$ $(\frac{7}{2},-\frac{1}{4})$ is a turning point (/extreme point/vertex).

$\begin{aligned} y & =f(x) \\ & = x^2-7x+12 \\ & = x^2-7x+\bigg( \frac{7}{2}\bigg)^2 - \bigg( \frac{7}{2}\bigg)^2 + 12 \\ & = \bigg( x-\frac{7}{2}\bigg)^2 - \frac{1}{4}\\ \end{aligned}$

iv. $\therefore$ The axis of symmetry of the graph is $x=\frac{7}{2}$ .

Differentiating $y=f(x)$ twice with respect to $x$ ,

$\begin{aligned} &\quad f''(x)\\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f'(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(2x-7) \\ & = 2 \quad (>0)\\ \end{aligned}$