To draw the graph of the given written in , we must start by identifying the values of $a,$ $b,$ and $c.$
$y=2x_{2}−6x+1⇔y=2x_{2}+(-6x)+1 $
We can see that $a=2,$ $b=-6,$ and $c=1.$ Now, we will follow four steps to graph the function.

- Find the .
- Calculate the .
- Identify the and its across the axis of symmetry.
- Connect the points with a .

### Finding the Axis of Symmetry

The axis of symmetry is a with equation

$x=-2ab .$ Since we already know the values of

$a$ and

$b,$ we can substitute them into the formula.

$x=-2ab $

$x=-2(2)-6 $

$x=1.5$

The axis of symmetry of the parabola is the vertical line with equation

$x=1.5.$ ### Calculating the Vertex

To calculate the vertex, we need to think of

$y$ as a function of

$x,$ $y=f(x).$ We can write the expression for the vertex by stating the

$x-$ and

$y-$coordinates in terms of

$a$ and

$b.$
$Vertex:(-2ab ,f(-2ab )) $
Note that the formula for the

$x-$coordinate is the same as the formula for the axis of symmetry, which is

$x=1.5.$ Thus, the

$x-$coordinate of the vertex is also

$1.5.$ To find the

$y-$coordinate, we need to substitute

$1.5$ for

$x$ in the given equation.

$y=2x_{2}−6x+1$

$y=2(1.5)_{2}−6(1.5)+1$

$y=2(2.25)−6(1.5)+1$

$y=4.5−9+1$

$y=-3.5$

We found the

$y-$coordinate, and now we know that the vertex is

$(1.5,-3.5).$ ### Identifying the $y-$intercept and its Reflection

The $y-$intercept of the graph of a quadratic function written in standard form is given by the value of $c.$ Thus, the point where our graph intercepts the $y-$axis is $(0,1).$ Let's plot this point and its reflection across the axis of symmetry.

### Connecting the Points

We can now draw the graph of the function. Since $a=2,$ which is positive, the parabola will open * upward *. Let's connect the three points with a smooth curve.