We have a , written in . To draw the graph of the related we must start by identifying the values of $a,$ $b,$ and $c.$
$y=x_{2}+3x+2⇔y=1x_{2}+3x+2 $
We can see that $a=1,$ $b=3,$ and $c=2.$ 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(1)3 $

$x=-1.5$

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.

$y=x_{2}+3x+2$

$y=(-1.5)_{2}+3(-1.5)+2$

$y=2.25+3(-1.5)+2$

$y=2.25−4.5+2$

$y=-0.25$

We found the

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

$(-1.5,-0.25).$ ### 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 $(0,2).$ 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 $a=1,$ which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.

By looking at the graph, we can state approximated values for the $x-$intercepts. We can see that the parabola intercepts the $x-$axis at $-2$ and $-1,$ approximately.