Pre-calculus Grade 11 IB (higher level)

In summary, the conversation discusses how to solve a question involving the transformation of a function and finding the values of a and b. The rules for transforming a function are given, and the resulting function g(x) is provided. The conversation then explains two methods for solving the problem, one involving working backwards and the other using equations. Finally, the values of a and b are found to be a=5 and b=7 by equating the coefficients of x and the free terms in the equations.
  • #1
TeddyJohnson
1
0
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.
 
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  • #2
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.
 
  • #3
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 $a=5,b=7$.
 

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