MHB Pre-calculus Grade 11 IB (higher level)

AI Thread Summary
To solve for a and b in the function transformation problem, the original function f(x) = ax + b undergoes a series of transformations resulting in g(x) = 4 - 15x. The transformations include a translation, reflection through the x-axis, and a horizontal stretch. By applying the transformation rules, the equations -3a = -15 and -a + b + 2 = 4 are derived, leading to the solutions a = 5 and b = 7. The discussion emphasizes understanding how each transformation affects the function's graph to arrive at the correct values of a and b.
TeddyJohnson
Messages
1
Reaction score
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.
 
Mathematics news on Phys.org
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.
 
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 $a=5,b=7$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top