Hong Kong

Stage 1 - Stage 3

Lesson

We've seen the form "$y=$`y`=" when we learnt about graphing. We've used this form to describe straight lines, parabolas and hyperbolas just to name a few.

In Is this a Functional Relationship, we were also introduced to the concept of functions. Where each input yielded a unique output.

When we are writing in function notation, instead of writing "$y=$`y`=", we write "$f(x)=$`f`(`x`)=". This gives us a bit more flexibility when we're working with equations or graphing as we don't have to keep track of so many $y$`y`s! Instead, using function notation, we can write $f(x)=$`f`(`x`)=, $g(x)=$`g`(`x`)=, $h(x)=$`h`(`x`)= and so on. These are all different expressions that involve only $x$`x` as the variable.

We can also evaluate "$f(x)$`f`(`x`)" by substituting values into the equations just like we would if the question was in the form "$y=$`y`=".

If $A(x)=x^2+1$`A`(`x`)=`x`2+1 and $Q(x)=x^2+9x$`Q`(`x`)=`x`2+9`x`, find:

**A)** $A(5)$`A`(5)

**Think:** This means we need to substitute $5$5 in for $x$`x` in the $A(x)$`A`(`x`) equation.

**Do:**

$A(5)$A(5) |
$=$= | $5^2+1$52+1 |

$=$= | $26$26 |

**B)** $Q(6)$`Q`(6)

**Think:** This means we need to substitute $6$6 in for $x$`x` in the $Q(x)$`Q`(`x`) equation.

**Do:**

$Q(6)$Q(6) |
$=$= | $6^2+9\times6$62+9×6 |

$=$= | $36+54$36+54 | |

$=$= | $90$90 |

**C)** $A(10)+Q(7)$`A`(10)+`Q`(7)

**Think**: This question will use both equations.

**Do:**

Let's work it out separately first:

$A(10)$A(10) |
$=$= | $10^2+1$102+1 |

$=$= | $101$101 | |

$Q(7)$Q(7) |
$=$= | $7^2+9\times7$72+9×7 |

$=$= | $112$112 | |

$101+112$101+112 | $=$= | $213$213 |

We can also do this as one long calculation:

$10^2+1+7^2+9\times7$102+1+72+9×7 | $=$= | $101+112$101+112 |

$=$= | $213$213 |

Consider the function $f\left(x\right)=8x+6$`f`(`x`)=8`x`+6.

Determine the output produced by the input value $x=5$

`x`=5.

If $Z(y)=y^2+12y+32$`Z`(`y`)=`y`2+12`y`+32, find $y$`y` when $Z(y)=-3$`Z`(`y`)=−3.

Write both solutions on the same line separated by a comma.