Chapter 4, Problem 17CR

To determine

To graph: The function $f\left(x\right)=2x^2-4x+1$ by determining whether the graph is concave up or

concave down and by finding the vertex, axis of symmetry, $y$ -intercepts and $x-$ intercepts

Explanation of Solution

Given information:

The function $f\left(x\right)=2x^2-4x+1$

Graph:

By comparing the function $f\left(x\right)=2x^2-4x+1$ with the standard form of the quadratic function $f\left(x\right)=a{x}^2+bx+c$ , $a=2,b=-4$ and $c=1$

The vertex of the quadratic function $f\left(x\right)=a{x}^2+bx+c$ is $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$ and the axis of symmetry is a vertical line $x=-\frac{b}{2a}$ .

  $x=-\left(\frac{-4}{2\left(2\right)}\right)=\frac{4}{4}=1$

Now, substitute $x=1$ in $f\left(x\right)=2x^2-4x+1$ .

  $f\left(1\right)=2{\left(1\right)}^2-4\left(1\right)+1=2-4+1=-1$

Thus, for the given quadratic function $f\left(x\right)=2x^2-4x+1$ vertex is $\left(1,-1\right)$

And the axis of symmetry is $x=1$

Since $a=2>0$ , so the graph of the function is concave up.

Now, determine the intercepts of the function $f\left(x\right)=2x^2-4x+1$

To determine the $x$ -intercepts, substitute $f\left(x\right)=0$ in the function $f\left(x\right)=2x^2-4x+1$ and solve for $x$

By substituting $f\left(x\right)=0$ in $f\left(x\right)=2x^2-4x+1$

  $0=2x^2-4x+1$

  $\Rightarrow 2x^2-3x-x+1=0$

Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to find the roots of the equation

By substituting the values of $a,b$ and $c$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(2\right)\left(1\right)}}{2\left(2\right)}$

  $\Rightarrow x=\frac{4\pm \sqrt{16-8}}{4}$

  $\Rightarrow x=\frac{4\pm \sqrt{8}}{4}$

  $\Rightarrow x=\frac{4\pm 2\sqrt{2}}{4}$

  $\Rightarrow x=1\pm \frac{\sqrt{2}}{2}$

Therefore, the $x$ - intercepts are $1+\frac{\sqrt{2}}{2}\approx 1.70711$ and $1-\frac{\sqrt{2}}{2}\approx 0.29289$

To determine the $y$ -intercepts, substitute $x=0$ in the equation $f\left(x\right)=2x^2-4x+1$ and solve for $y$

By substituting $x=0$ in $f\left(x\right)=2x^2-4x+1$

  $f\left(x\right)=2{\left(0\right)}^2-4\left(0\right)+1$

By simplifying,

  $\Rightarrow f\left(x\right)=0-0+1$

  $\Rightarrow f\left(x\right)=1$

Therefore, the $y$ - intercept is $1$

To get the additional point, use the axis of symmetry line $x=1$ and the $y-$ intercept $\left(0,1\right)$

Substitute $y=1$ in the function $f\left(x\right)=2x^2-4x+1$

  $1=2x^2-4x+1$

  $\Rightarrow 0=2x^2-4x$

  $\Rightarrow 2x^2-4x=0$

  $\Rightarrow x\left(2x-4\right)=0$ .

  $\Rightarrow x=0$ or $2x-4=0$

  $\Rightarrow x=0$ or $2x=4$

  $\Rightarrow x=0$ or $x=2$

Thus, the additional point is $\left(2,1\right)$

To draw the graph, plot the vertex $\left(1,-1\right)$ , axis of symmetry $x=1$ , and intercepts $\left(1.70711,0\right),\left(0.29289,0\right)$ and $\left(0,1\right)$ , additional point $\left(2,1\right)$ of the quadratic function to draw the graph of the function $f\left(x\right)=2x^2-4x+1$ .

The graph is as follows:

  

Interpretation:

The graph of the function $f\left(x\right)=2x^2-4x+1$ represents an opening upward parabola.