Question

Graph $f\left(x\right)=2x^2-4x+1$by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any.

Short answer

The graph can be given as

Step-by-step solution

Given information

We are given a function $f\left(x\right)=2x^2-4x+1$

As the coefficient of $x^2$is positive the parabola opens upward.

We find the intercepts

At x-intercept $f\left(x\right)=0$

$$\begin{gathered} 2x^2-4x+1=0 \\ x=\frac{4\pm \sqrt{16-4\left(2\right)}}{4} \\ x=\frac{4\pm \sqrt{16-8}}{4} \\ x=\frac{4\pm \sqrt{8}}{4} \\ x=0.293orx=1.707 \end{gathered}$$

At y-intercept $x=0$

We have,

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

Hence the intercepts are$(0,1)(1.707,0),(0.293,0)$

Find the vertex

The vertex can be given as

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{-4}{4} \\ x=1 \end{gathered}$$

Hence, $$\begin{gathered} f\left(x\right)=2x^2-4x+1 \\ f\left(1\right)=2-4+1 \\ f\left(1\right)=-1 \end{gathered}$$

The coordinate of vertex is$(1,-1)$

Plot the graph

The graph can be given as