91Ó°ÊÓ

Consider the quadratic function $f\left(x\right)=2x^2-4x+2$.

For this parabola, the graph will open upwards as the leading coefficient is positive.

Then finding the axis of symmetry, vertex, the intercepts, and a point with help of the axis of symmetry, will help us to plot the points of the graph and get the required graph.

On comparing the given function $f\left(x\right)=2x^2-4x+2$with the standard form $f\left(x\right)=a{x}^2+bx+c$we get

$$\begin{gathered} a=2 \\ b=-4 \\ c=2 \end{gathered}$$

The axis of symmetry for the standard form is $x=-\frac{b}{2a}$.

So for the given function, the axis of symmetry is given as

$$\begin{gathered} x=-\frac{-4}{2·2} \\ x=-(-1) \\ x=1 \end{gathered}$$

Thus the equation of the axis of symmetry is $x=1$.

Vertex is a point on the line of axis of symmetry.

Here axis of symmetry is $x=1$.

So the function value $f\left(1\right)$is given as

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

So the vertex of the parabola is $(1,0)$.

To find the y-intercept substitute $0$for $x$in the function.

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

Thus the y intercept is $(0,2)$.

To find $x$intercepts substitute $0$for $f\left(x\right)$in the function.

$\begin{array}{rcl}f\left(x\right)& =& 2x^2-4x+2\\ 0& =& 2x^2-4x+2\end{array}$

Factor the perfect square trinomial.

$$\begin{gathered} 0=2(x^2-2x+1) \\ 0=x^2-2x+1 \\ 0={(x-1)}^2 \end{gathered}$$

On solving

role="math" localid="1645158554179" $\begin{array}{rcl}x-1& =& 0\\ x& =& 1\end{array}$

So the x intercept is $(1,0)$.

We havey intercept as $(0,2)$.

The axis of symmetry is $x=1$.
It means that the intercept is at a distance of one unit to the left of the axis of symmetry.
We can get a point on the graph using this information.
A point at the same distance, i.e., one unit, but to the right will also lie on the graph. This point will be $(2,2)$.

Plotting all the points we got and joining these points will give us the graph as-