# Math Help - h(31)=?

1. ## h(31)=?

"The period of function h is 15. If h (1) = 0 and h(-6) = 3, calculate the value of:

a) h(31)
b) h(9)
c) h(-14)
d) d(-51)"

I just need to see the work for a) at least to understand what this problem is all about so please show me how to do a). Thanks in advance!

2. Hello, s3a!

You're expected to know what a periodic function is.

The period of function $h$ is 15.

If $h (1) = 0$ and $h(\text{-}6) = 3$, calculate the values of:

. . $(a)\;h(31) \qquad(b)\;h(9)\qquad(c)\;h(\text{-}14)\qquad(d)\;h(\text{-}51)$

The function is periodic ... and "repeats itself" every 15 units.

$\text{So if }h(1) = 0\text{, then: }\;h(16) = 0,\;\;\underbrace{h(31) = 0}_{(a)},\;\;h(46) = 0,\;\; \hdots$

$\text{Moving "backwards": }\;h(1) = 0,\;\;\underbrace{h(\text{-}14) = 0}_{(c)},\;\;h(\text{-}29)=0,\;\;\hdots$

Get the idea?

3. Originally Posted by Soroban
Hello, s3a!

You're expected to know what a periodic function is.

The function is periodic ... and "repeats itself" every 15 units.

$\text{So if }h(1) = 0\text{, then: }\;h(16) = 0,\;\;\underbrace{h(31) = 0}_{(a)},\;\;h(46) = 0,\;\; \hdots$

$\text{Moving "backwards": }\;h(1) = 0,\;\;\underbrace{h(\text{-}14) = 0}_{(c)},\;\;h(\text{-}29)=0,\;\;\hdots$

Get the idea?

-14+15=1 not 0 for the backward one. I get the one on top though.

4. Yes: ${\color{blue}-14 + 15} = 1$

That is why: $h(-14) = h \Big({\color{blue}-14 + 15} \Big) = h(1) = 0$