Question

How do you find the axis of symmetry, and the maximum or minimum value of the function `f(x)=x^2 -2x -15`?

Answer 1

Axis of symmetry `x=1`
Minimum value `=-16`

Explanation:

The parabola opens upward and so this function has a minimum value.
To solve for the minimum value we solve for the vertex.

`y=ax^2+bx+c`

`y=1*x^2+(-2)*x+(-15)`

so that `a=1` and `b=-2` and `c=-15`

Vertex `(h, k)`

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

`h=(-(-2))/(2(1))=1`

`k=c-b^2/(4a)`

`k=-15-(-2)^2/(4(1))`

`k=-15-1`

`k=-16`

Vertex `(h, k)=(1, -16)`

The minimum value of the function is `f(1)=-16`

Kindly see the graph of `f(x)=x^2-2x-15` with the axis of symmetry `x=1` dividing the parabola into two equal parts.
graph{(y-x^2+2x+15)(y+1000x-1000)=0[-36,36,-18,18]}

God bless ....I hope the explanation is useful.

Answer 2

Axis of symetry `x=1`
Value of the function `y=-16`

Explanation:

Given -

`y=x^2-2x-15`

Find Axis of symetry.

`x=(-2b)/(2a)=(-(-2))/(2 xx 1)=2/2=1`

Axis of symetry `x=1`

Maximum of Minimum Values

`dy/dx=2x-2`
`(d^2y)/(dx^2)=2`
`dy/dx=0 =>2x-2=0`
`x=2/2=1`

At `(x=1): dy/dx=0;(d^2y)/(dx^2)>0`
Hence there is a minimum at `x=1`

Value of the function

`y=1^2-2(1)-15`
`y=1-2-15=-16`