Chapter 2.1, Problem 26E

To determine

To describe:The graph of given quadraticfunction and identify the vertex, and verify the results with graphing utility.

Explanation of Solution

Given data: $y$

The given quadratic function is $f\left(x\right)=x^2+3x+\frac{1}{4}$ .

Concept used:

The standard form of quadratic function $f\left(x\right)=a{\left(x-\text{α}\right)}^2+\text{β}$ represents a parabola with vertex is $\left(\text{α,β}\right)$ . If $a>0$ then the parabola has minima at the vertex, so concave upwards, and if $a<0$ then the parabola has maxima at the vertex so concave downwards

Calculations:

The given quadratic function is $f\left(x\right)=x^2+3x+\frac{1}{4}$ may be written as

  $f\left(x\right)=x^2+\cdot 2\cdot \frac{3}{2}\cdot x+\frac{9}{4}-2$

  $\Rightarrow f\left(x\right)=1\cdot {\left(x-\frac{3}{2}\right)}^2-2$............. (1)

Now, comparing equation (1) with the function $f(x)=a{\left(x-\text{α}\right)}^{\text{2}}\text{+β}$ , we have $a\text{=}1$ , $h\text{=}\frac{3}{2}$ , and $k=-2$ .

Thus, the quadratic function $f\left(x\right)=x^2+3x+\frac{1}{4}$ represents a parabola with vertex $\left(\frac{3}{2},-2\right)$ .

Since, $a=\text{1>0}$ , hence the parabola is minimum at the vertex so is concave upwards.

The graph of the function $f\left(x\right)=x^2+3x+\frac{1}{4}$ (a parabola) is shown in Figure-1 here.

We observe that the graph of function $f\left(x\right)=x^2+3x+\frac{1}{4}$ is an upwards concave parabola with vertex $\left(\frac{3}{2},-2\right)$ that verify the results described above.

  

The graph of the function $f\left(x\right)=x^2+3x+\frac{1}{4}$ is obtained by translatingthe graph of function $g\left(x\right)=x^2$ , (i) through a distance 3/2 unitalong negative x −axis, and then 2 unit along negative y -axis.