Chapter 2.1, Problem 18E

To determine

To Write the quadratic function in standard form and sketch its graph.

Identify the vertex, axis of symmetry and x-intercept.

Explanation of Solution

Given:

The function is $f(x)=x^2+16x+61$ .

Formula used:

1. The x coordinate of vertex and the axis of symmetry can be given as,

  $x=\frac{-b}{2a}\mathrm{...}\left(1\right)$

2. The x-intercept can be calculated by substituting y value zero.

Calculation:

Consider, $f(x)=x^2+16x+61\mathrm{....}(2)$

Standard form of quadratic function is given as,

  $y=a{x}^2+bx+c....(3)$

Comparing the coefficient of equation (2) and (3),

  $a=1,b=16\text{and}c=61$

The x coordinate of vertex can be given as,

  $\begin{array}{c}x=\frac{-16}{2\times 1}\\ =-8\end{array}$

Substitute the value of x in the function,

  $\begin{array}{c}f(x)={(}^{-}+16\left(-8\right)+61\\ =64-128+61\\ =-3\end{array}$

So, the axis of symmetry can be calculated as,

  $x=-8$

Substitute $y=0$ in the equation of function,

  $\begin{array}{l}0=x^2+16x+61\\ x=\frac{-16\pm \sqrt{{16}^2-4\times 1\times 61}}{2\times 1}\\ x=\frac{-16\pm 3.464}{2}\\ x=-9.732,-6.268\end{array}$

The graph of the above equations is given below,

  

Therefore, the vertex, axis of symmetry and x-intercept are $(-8,-3)$ , $x=-8$ and $\left(-6.268,0\right),(-9.732,0)$ respectively.