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Pre-calculus Grade 11 IB (higher level)

TeddyJohnson

New member
Dec 2, 2018
1
Can anyone explain how to solve this question, please? The answer is a=5 & b=7, but I don't understand how to solve it.

The graph of function f(x) = ax + b is transformed by the following sequence:

translation by (1) (meaning 1 horizontal, 2 vertical)
(2)

reflection through y=0

horizontal stretch, scale factor 1/3, relative to x=0

The resulting function is g(x)=4-15x
Find a & b

Thanks for your help.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
Hi, and welcome to the forum!

To solve this problem you need to know how change of function $f$ affects the graph of $f$. Here are a few rules.
  1. $f(x)\mapsto f(x-a)$: horizontal shift by $a$ to the right.
  2. $f(x)\mapsto f(x)+b$: vertical shift by $b$ up.
  3. $f(x)\mapsto f(-x)$: reflection through $y=0$.
  4. $f(x)\mapsto f(x/k)$: horizontal stretch with scale factor $k$ relative to $x=0$.
Suppose the original function is $f(x)=ax+b$. Using these rules, the formula changes as follows.
  1. Translation by (1, 2): $a(x-1)+b+2$.
  2. Reflection through $y=0$: $a(-x-1)+b+2$.
  3. Horizontal stretch with scale factor $1/3$ relative to $x=0$: $a(-3x-1)+b+2$.
The problem statement says that $a(-3x-1)+b+2=4-15x$. Equating the numbers multiplied by $x$ and the free coefficient we get two equations: $-3a=-15$ and $-a+b+2=4$, from where $a=5$ and $b=7$.

It is important to remember that when viewing a formula like $a(-x-1)+b+2$ as a function of $x$, only $x$ changes when we go, say, from $f(x)$ to $f(3x)$. The result is $a(-(3x)-1)+b+2$ and not $3(a(-x-1)+b+2)$.

Here is another way. The points in the original graph have coordinates $(x, ax+b)$. The geometric transformation change the coordinates as follows.
  1. Translation by (1, 2): $(x+1,ax+b+2)$.
  2. Reflection through $y=0$: $(-(x+1),ax+b+2)$.
  3. Horizontal stretch with scale factor $1/3$ relative to $x=0$: $(-(x+1)/3,ax+b+2)$.
The resulting point is $(x', 4-15x')$ for some $x'$. Therefore $x'=-(x+1)/3$ and $4-15x'=4+5(x+1)=5x+9$. Equating this with $ax+b+2$ (separately coefficients at $x$ and the free one) we get $a=5$ and $b+2=9$, i.e., $b=7$.

If you need more explanation, feel free to ask.
 

Olinguito

Well-known member
Apr 22, 2018
251
I think there’s a typo: the reflection should be through $\color{red}x\color{black}=0$, not $y=0$.

Here’s yet another way: work backwards.

Start with $g(x)=4-15x$.

Do the reverse of horizontal stretching by $\frac13$, namely horizontal stretching by $3$. Under this mapping, $(x',y')=(3x,y)$ $\implies$ $(x,y)=\left(\frac13x',y'\right)$ $\implies$ $g(x)=4-15x\mapsto h_1(x)=4-15\left(\frac13x\right)=4-5x$.

Next, the reverse of reflection in $x=0$, which is the same transformation: $(x',y')=(-x,y)$ $\implies$ $(x,y)=\left(-x',y'\right)$ $\implies$ $h_1(x)=4-5x\mapsto h_2(x)=4-5(-x)=4+5x$.

Finally, the reverse of the translation $\begin{pmatrix}1 \\ 2\end{pmatrix}$, which is $\begin{pmatrix}-1 \\ -2\end{pmatrix}$: $(x',y')=(x-1,y-2)$ $\implies$ $(x,y)=\left(x'+1,y'+2'\right)$ $\implies$ $h_2(x)=4+5x\mapsto f(x)+2=4+5(x+1)=9+5x$, i.e. $f(x)=5x+7$.

Hence $a=5,b=7$.