Chapter 8, Problem 24CR

To determine

Tograph:The quadratic function, label its vertex, axis of symmetry, and x -intercept. Also to describe the domain and range of the function.

Answer to Problem 24CR

The domain of function $y=x^2-8x+15$ is set of all real numbers $D=ℝ$ or $D=\left(-\infty ,\infty \right)$ . The range of $y=x^2-8x+15$ is $\left[-1,\infty \right)$ .

Explanation of Solution

Given information:

The given function is $y=x^2-8x+15$ .

Concept /Formula/ Method used:

The graph of function is drawn using graphing calculator. The vertex of quadratic function point of its maximum or minimum. The axis of symmetry is the normal to its tangent at vertex and x -intercept is the point where its graph intersect x -axis. Domain is the set of variables x on which on which function is defined and the range is the set of images of members of the domain.

Calculations:

The given function is $y=x^2-8x+15$

Or

  $y=x^2-8x+15=\left(x-3\right)\left(x-5\right)$

Or

  $y=x^2-8x+16-1={\left(x-4\right)}^2-1$

The standard form of quadratic function is $y=a{\left(x-h\right)}^2+k$ . The vertex is $\left(h,k\right)$

The general form of the quadratic function is $y=a{x}^2+bx+c$ . The axis of is $x=-\frac{b}{2a}$ .

The x -intercepts of the quadratic function are points where $y=0$ .

The vertex of the function $y=x^2-8x+15$ is $\left(4,-1\right)$ .

The axis of symmetry of the function $y=x^2-8x+15$ is $x=-\frac{-(-8)}{2\times 1}=4$ .

Since, $y=\left(x-3\right)\left(x-5\right)=0\Rightarrow x=3,5$

The x -intercepts of the function $y=x^2-8x+15$ are $\left(3,0\right)$ and $\left(5,0\right)$ .

The graph of function $y=x^2-8x+15$ with labeled vertex $\left(4,-1\right)$ , axis of symmetry $x=4$ , and x -intercepts $\left(3,0\right)$ and $\left(5,0\right)$ is shown in Figure-1 here.

Domain:

Since the function $y=x^2-8x+15$ is defined for all real numbers so its domain D is the set of all real numbers $D=ℝ$ or $D=\left(-\infty ,\infty \right)$ .

Range:

The function is $y={\left(x-4\right)}^2-1$

  $\Rightarrow y+1={\left(x-4\right)}^2$

  $\Rightarrow x=4+\sqrt{y+1}$

For x to be real $y+1>0$

  $\Rightarrow y\ge -1$

This shows that the range of $y=x^2-8x+15$ is $\left[-1,\infty \right)$ .

Conclusion:

The domain of function $y=x^2-8x+15$ is set of all real numbers $D=ℝ$ or $D=\left(-\infty ,\infty \right)$ . The range of $y=x^2-8x+15$ is $\left[-1,\infty \right)$ .