How do you graph the quadratic function and identify the vertex and axis of symmetry and x intercepts for `y=1/3(x+4)(x+1)`?

Given:

`y=1/3(x+4)(x+1)`

Expand `(x+4)(x+1)`.

`y=1/3(x^2+5x+4)`

Distribute `1/3`.

`y=1/3x^2+5/3x+4/3` is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=1/3`, `b=5/3`, and `c=4/3`

The vertex is the maximum or minimum point of the parabola. The formula for the axis of symmetry gives us the x-coordinate of the vertex:

`x=(-b)/(2a)`

`x=(-5/3)/(2*1/3)`

`x=(-5/3)/(2/3)`

`x=-5/3xx3/2`

`x=-15/6`

`x=-5/2` or `2.5`

To find the y-coordinate of the vertex, substitute `-5/2` for `x` and solve for `y`.

`y=1/3(-5/2)^2+5/3(-5/2)+4/3`

`y=1/3(25/4)-25/6+4/3`

`y=25/12-25/6+4/3`

The least common denominator is `12`. Multiply `25/6xx2/2` and `4/3xx4/4` to get equivalent fractions. Since `n/n=1`, the numbers will change but the value of each fraction will not change.

`y=25/12-25/6xxcolor(red)2/color(red)2+4/3xxcolor(blue)4/color(blue)4`

`y=25/12-50/12+16/12`

`y=-9/12`

`y=-3/4` or `-0.75`

The vertex is `(-5/2,-3/4)` or `(-2.5,-0.75)`. Plot this point.

The y-intercept is the value of `y` when `x=0`. Substitute `0` for `x` and solve for `y`.

`y=1/3(0)^2+5/3(0)+4/3`

`y=4/3` or `~~1.333`

The y-intercept is `(0,4/3)` or `(0,~~1.333)`. Plot this point.

The x-intercepts are the values for `x` when `y=0`. Substitute `0` for `y` and solve for `x`.

`0=1/3x^2+5/3x+4/3`

Switch sides.

`1/3x^2+5/3x+4/3=0`

Multiply both sides by `3`.

`x^2+5x+4=0`

Factor `x^2+5x+4`.

`(x+1)(x+4)=0`

Set each binomial to zero and solve.

`x+1=0`

`x=-1`

`x+4=0`

`x=-4`

The x-intercepts are `(-1,0), (-4,0)`. Plot these points.

Additional point: `x=-5`

`x=-5` is the mirror of the x-coordinate of the y-intercept.

Substitute `-5` for `x` and solve for `y`.

`y=1/3(-5)^2+5/3(-5)+4/3`

`y=25/3-25/3+4/3`

Additional point: `(-5,4/3)` or `(-5,~~1.333)`. Plot this point.

Sketch a graph through the points. Do not connect the dots.

graph{y=(x^2)/3+(5x)/3+4/3 [-10, 10, -5, 5]}