Question

How do you graph, label the vertex, axis of symmetry and x intercepts of `y=(3x-1)(x-3)`?

Answer 1

graph{(3x-1)(x-3) [-20, 20, -10, 10]}
Vertex = ( `5/3`, `-8/3`)
Axis of symmetry = `x=5/3`
X-intercepts = `(1/3,0) and (3,0)`

Explanation:

Multiplying the two brackets we get the quadratic,
`3x^2-10x+3`
Comparing with `ax^2+bx+c`, we get
`a=3 , b=-10, c=3`
and the Discriminant = `b^2-4ac` => `64`
Co-ordinates of Vertex of parabola are (`-b/(2a)`,`-D/(4a)`)
plug the the values of `a,b,c` to get vertex of parabola.

Axis of symmetry is the x-coordinate of Vertex i.e
`x=5/3`

x-intercept of the parabola is basically the roots of equation.
Roots can be obtained by equating the function to `0`.
`=>(3x-1)(x-3)=0`
either `3x-1=0` or `x-3=0` (By Zero Product Rule)
this gives `x=1/3 , 3`
These are the x-intercepts.

Answer 2

X-intercept

`x=1/3`
`x=3`
Vertex `(5/3, -16/3)`
Axis of symmetry `x=5/3`

Explanation:

Given -

`y=(3x-1)(x-3)`

X-intercept
At `y=0`

`(3x-1)(x-3)=0`

`3x=1`
`x=1/3`

`x=3`

`y=3x^2-x-9x+3`
`y=3x^2-10x+3`

Vertex

`x=(-b)/(2a)=(-(-10))/(2 xx3)=10/6=5/3`

At `x=5/3`

`y=3(5/3)^2-10(5/3)+3`
`y=3(25/9)-50/3+3`
`y=25/3-50/3+3`
`y=(25-50+9)/3=-16/3`

Vertex `(5/3, -16/3)`

Axis of symmetry `x=5/3`