Chapter 7.6, Problem 29WE

a.

To determine

The vertex of the graph of the function $f\left(x\right)=x^2-2x-5$ .

Answer to Problem 29WE

The vertex of graph of function is $\left(1,-6\right)$ .

Explanation of Solution

Given information: The given function is $f\left(x\right)=x^2-2x-5$ .

Calculation: The given function can be written as,

  $\begin{array}{c}y=x^2-2x-\mathrm{5....}\left(\text{Replace }f\left(x\right)=y\right)\\ \text{or, }y+5=x^2-2x\\ \text{or, }y+5+\left(1\right)=x^2-2\left(x\right)\left(1\right)+{\left(1\right)}^2\left(\text{Complete square}\right)\\ y+6={\left(x-1\right)}^2\mathrm{....}\left(1\right)\end{array}$

Since (1) represents a parabola of the form $y-k=a{\left(x-h\right)}^2$ . Therefore the graph for the function has vertex $\left(1,-6\right)$ .

b.

To determine

The domain of the function $f\left(x\right)=x^2-2x-5$ .

Answer to Problem 29WE

The domain of the function is all real numbers.

Explanation of Solution

Given information: The given function is $f\left(x\right)=x^2-2x-5$ .

Calculation: From (1), the parabola obtained represents a parabola opening upwards because of positive coefficient of ${\left(x-1\right)}^2$ . Therefore the domain of the function has no restrictions at all.

Hence the domain of the function is all real numbers.

c.

To determine

The range of the function $f\left(x\right)=x^2-2x-5$ .

Answer to Problem 29WE

The range of the function is $\left\{y|y\ge -6\right\}$ .

Explanation of Solution

Given information: The given function is $f\left(x\right)=x^2-2x-5$ .

Calculation: From (1) the parabola obtained represents a parabola opening upwards because of positive coefficient of ${\left(x-1\right)}^2$ . Also the parabola has the vertex $\left(1,-6\right)$ which means the lower limit for range is $-6$ . Since, the parabola (1) is lower bound, therefore the range of the function is all real numbers greater than or equal to $-6$ . Mathematically, the range of function is $\left\{y|y\ge -6\right\}$ .

d.

To determine

The zeros of the function $f\left(x\right)=x^2-2x-5$ .

Answer to Problem 29WE

The zeros of the function are $1+\sqrt{6},1-\sqrt{6}$ .

Explanation of Solution

Given information: The given function is $f\left(x\right)=x^2-2x-5$ .

Calculation: The zeros of the function can be obtained by equating the given function equal to zero. Therefore $f\left(x\right)=0$ gives,

  $\begin{array}{c}{x}^2-2x-5=0\\ x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-5\right)}}{2\left(1\right)}\left(\text{Quadratic formula}\right)\\ x=\frac{2\pm 2\sqrt{6}}{2}\\ x=1+\sqrt{6},1-\sqrt{6}\end{array}$

Hence the zeros of the function are $1+\sqrt{6},1-\sqrt{6}$ .