Chapter 2.1, Problem 30E

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 $f\left(x\right)=2x^2-x+1$ .

Concept used:

The standard form of quadratic function $h\left(x\right)=a{\left(x-\text{α}\right)}^2+\text{β}$ 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 $f\left(x\right)=2x^2-x+1$ may be written as

  $f\left(x\right)=2\left(x^2-2.\frac{1}{4}x+\frac{1}{16}\right)+\frac{7}{8}$

  $\Rightarrow f\left(x\right)=2{\left(x-\frac{1}{4}\right)}^2+\frac{7}{8}$............(1)

Now, comparing equation (1) with the function $h(x)=a{\left(x-\text{α}\right)}^{\text{2}}\text{+β}$ , we have $a\text{=}2$ , $h\text{=}\frac{1}{4}$ , and $k=\frac{7}{8}$ .

Thus, the quadratic function $f\left(x\right)=2x^2-x+1$ represents a parabola with vertex $\left(\frac{1}{4},\frac{7}{8}\right)$ .

Since, $a=4>0$ , hence the parabola has minimum at the vertex so is concave upwards.

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

We observe that the graph of function $f\left(x\right)=2x^2-x+1$ is an upwards concave parabola with vertex $\left(\frac{1}{4},\frac{7}{8}\right)$ that verify the results described above.

  

The graph of the function $f\left(x\right)=2x^2-x+1$ is by(i) shifting the graph of function $g\left(x\right)=x^2$ through a distance $\frac{7}{8}$ units along y -axis, then (ii) shifting it further through distance $\frac{1}{4}$ unit alongx −axis.