Question

Convert the equation `f(t)=t^2+12t-18` into vertex form and hence find the minimum value of `f(t)`?

Answer 1

The function is minimum at vertex `(-6,-54)`

Explanation:

Part A - The equation `f(t)=t^2+12t-18` can be written as

`t^2+2xx6xxt+6^2-36-18`

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

This is the equation in vertex form, which is of the type

`y=a(x-h)^2+k` or `(y-k)=a(x-h)^2`, where `(h,k)` is the vertex.

Part B - Hence, vertex of `t^2+12t-18` is `(-6,-54)`

Observe that at `t=-6`, `f(t)=-54`, when `(x-6)^2=0`.

As `t` increases or decreases i.e. moves on either side from `-6`, `(x+6)^2` can only increase (as it is always positive).

Hence at vertex `t^2+12t-18` is minimum.

graph{x^2+12x-18 [-45.17, 34.83, -55.52, -15.52]}