Question

The function f(t)=t^2+12t-18 represents a parabolaPart A: rewrite the function in vertex formPart B: determine the vertex and indicate whether it is a maximum or a minimum on the graphPart C: determine the axis of symmetry for f(t)

Answer 1

A general form o quadratic function is given by the following expression:

`f(x)=ax^2+bx+c`

The vertex form of this quadratic function is given by the following expression:

`f(x)=a(x-h)^2+k`

where (h,k) is the vertex of the quadratic function.

PART A:

To find the vertex form of the given function, we must complete the square of this function:

`t^2+12t-18+6^2-6^2`

this is equivalent to:

`(t+6)^2-18-6^2`

that is:

`(t+6)^2-54`

so that, the vertex form of the given function is:

`f(t)=(t+6)^2-54`

Part B: According to the previous part, we can conclude that the vertex is:

`(h,k)=(-6,-54)`

Part C: The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola. So that, we can conclude that the axis of symmetry for f(t) is:

`x\text{ =-6}`