Question

How do you find the vertex of the parabola `y = x^2 - 4x`?

Answer 1

`(2, -4)`

Explanation:

The easiest way is:

`y=ax^2+bx +c`

axis on symmetry is:

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

Vertex is: `(aos, f(aos))`

c = y-intercept

so your function:

`y = x^2 - 4x`

a = 1
b = -4
c = 0

`aos = (-(-4))/(2*1) = 2`

f(aos) means we put the aos back in your function as x and solve for y:

`f(aos) = f(2) = 2^2 - 4*2 = -4`

Vertex is: `(aos, f(aos))`

Vertex is: `(2, -4)`

Note, this can also be solved by completing the square and converting the function to vertex form.

graph{y = x^2 - 4x [-7.13, 12.87, -7.8, 2.2]}

Answer 2

`"vertex "=(2,-4)`

Explanation:

`"the vertex lies on the axis of symmetry which is"`
`"situated at the midpoint of the zeros"`

`"to find the zeros set y = 0"`

`rArrx^2-4x=0larrcolor(blue)"factorise"`

`rArrx(x-4)=0`

`"equate each factor to zero and solve for x"`

`rArrx=0`

`x-4=0rArrx=4`

`"axis of symmetry/x-coordinate of vertex "=(0+4)/2=2`

`"substitute this value into the equation for y"`

`rArry=2^2-(4xx2)=4-8=-4`

`rArrcolor(magenta)"vertex "=(2,-4)`
graph{(y-x^2+4x)((x-2)^2+(y+4)^2-0.04)=0 [-10, 10, -5, 5]}