## Calculus, 10th Edition (Anton)

(a)$$x_1=4, \qquad x_2=2$$ (b) $y$-coordinate of no point of the parabola is $10$. (c)$$x \in \mathbb R -(2,4)$$ (d) This function has no maximum. $(3,-1)$ is the minimum point of this function.
(a) Let us find the values of $x$ for which $x^2-6x+8=0$.$$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\quad a=1, \, b=-6, \, c=8 \\ \Rightarrow \quad x_{1,2}=\frac{6 \pm \sqrt{36-32}}{2} \quad \Rightarrow \quad x_1=4, \, x_2=2$$ (b) Let us find the values of $x$ for which $x^2-6x+8=-10$.$$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\quad a=1, \, b=-6, \, c=18$$The above equation has no root since$$b^2-4ac=36-4 \cdot 18<0.$$Thus, $y$-coordinate of no point of the parabola is $10$. (c) Let us find the values of $x$ for which $x^2-6x+8 \ge 0.$ $$\Rightarrow \quad (x-4)(x-2) \ge 0 \\ \Rightarrow \quad x \ge 4, \, x \ge 2 \quad \text{ or } \quad x \le 4, \, x \le 2 \\ x \in \mathbb R- (2,4).$$ (d) $y=x^2-6x+8=(x-4)(x-2)$ has no maximum since for example, for $x_1, x_2 \ge 4$ and $x_1 < x_2$, we have $y_1=(x_1-4)(x_1-2) > (x_2-4)(x_2-2)$, so there is no upper bound for $y=x^2-6x+8$. However, $y=x^2-6x+8$ has minimum since it represents a parabola. Since the minimum is the midpoint of any two points on the parabola having the same $y$-coordinate, we can find the $x$-coordinate of the minimum point by means of averaging the $x$-coordinate of roots ($y=0)$:$$x_{\text{min}}=\frac{x_1+x_2}{2}= \frac{2+4}{2}=3$$For $x_{\text{min}}=3$ we have $y_{\text{min}}=3^2-6 \cdot 3 +8=-1$. Thus, $(3, -1)$ is the minimum point of the parabola.