Chapter 2.1, Problem 22E

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 $f(x)=x^2+3x+\frac{1}{4}$ .

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, $f(x)=x^2+3x+\frac{1}{4}....(2)$

Standard form of quadratic function is given as,

  $y=a{x}^2+bx+c....(3)$

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

  $a=1,b=3\text{and}c=\frac{1}{4}$

The x coordinate of vertex can be given as,

  $\begin{array}{c}x=\frac{-3}{2\times 1}\\ =\frac{-3}{2}\end{array}$

Substitute the value of x in the function,

  $\begin{array}{c}f(x)={\left(\frac{-3}{2}\right)}^2+3\times (\frac{-3}{2})+\frac{1}{4}\\ =\frac{9}{4}-\frac{9}{2}+\frac{1}{4}\\ =-2\end{array}$

So, the axis of symmetry can be calculated as,

  $x=\frac{-3}{2}$

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

  $\begin{array}{l}0=x^2+3x+\frac{1}{4}\\ x=\frac{-3\pm \sqrt{{(3)}^2-4\times 1\times \frac{1}{4}}}{2\times 1}\\ x=\frac{-3\pm \sqrt{8}}{2}\\ x=-2.914,-0.086\end{array}$

The graph of the above equations is given below,

  

Therefore, the vertex, axis of symmetry and x-intercept are $(-1.5,-2)$ , $x=-1.5$ and $\left(-2.914,0\right),\left(-0.086,0\right)$ respectively.