Chapter 2.1, Problem 26E

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

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, $f(x)=2x^2-x+1\mathrm{....}(2)$

Standard form of quadratic function is given as,

  $y=a{x}^2+bx+c....(3)$

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

  $a=2,b=-1\text{and}c=1$

The x coordinate of vertex can be given as,

  $\begin{array}{c}x=\frac{-(-1)}{2\times 2}\\ =\frac{1}{4}\end{array}$

Substitute the value of x in the function,

  $\begin{array}{c}f(x)=2{(}^{\frac{1}{4}}-\left(\frac{1}{4}\right)+1\\ =\frac{1}{8}-\frac{1}{4}+1\\ =\frac{7}{8}\end{array}$

So, the axis of symmetry can be calculated as,

  $x=\frac{1}{4}$

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

  $\begin{array}{l}0=2x^2-x+1\\ x=\frac{1\pm \sqrt{{(-1)}^2-4\times 2\times 1}}{2\times 2}\\ x=\frac{1\pm \sqrt{-7}}{4}\end{array}$

The square root of a negative number is imaginary. This suggests that this equation does nothave an x-intercept

The graph of the above equations is given below,

  

The graph can verify what’s going on. Notice that the graph crossed the y-axis at (0, 1), but never did with the x-axis

Therefore, the vertex andaxis of symmetryare $(\frac{1}{4},\frac{7}{8})$ and $x=\frac{1}{4}$ respectively.