IB Maths AA HL Transformations and Modulus Questions

1. Interpret the translation vector $(\genfrac{}{}{0ex}{}{-2}{3})$: this shifts the graph $2$ units to the left and $3$ units up.

2. Apply the transformations to the function notation: $g(x)=f(x-(-2))+3=f(x+2)+3$.

3. Substitute $(x+2)$ into the original function: $g(x)=(x+2{)}^2-4+3$.

4. Expand the binomial square: $g(x)=(x^2+4x+4)-4+3$.

5. Simplify the constant terms: $g(x)=x^2+4x+4-1$.

6. State the final expanded expression: $\mathbf{g}\mathbf{(}\mathbf{x}\mathbf{)}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}$.

1. Set up the equation: $|2x-6|=x$.

2. Form Case 1, assuming the argument is positive (so the modulus does nothing): $2x-6=x$.

3. Solve Case 1: Subtract $x$ and add $6$ to get $x=6$.

4. Form Case 2, assuming the argument is negative (so the modulus multiplies it by $-1$): $-(2x-6)=x⟹-2x+6=x$.

5. Solve Case 2: Add $2x$ to get $3x=6⟹x=2$.

6. Verify both solutions in the original equation to check for extraneous roots. For $x=6$: $|12-6|=6$ (Valid). For $x=2$: $|4-6|=|-2|=2$ (Valid). Solutions: $\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{6}$.

1. Factorise the argument of the trigonometric function so the coefficient of $x$ is separated: $2x-\pi =2(x-\frac{\pi }{2})$.

2. Identify the horizontal stretch from the multiplier $2$: A horizontal stretch by scale factor $\frac{1}{2}$.

3. Identify the horizontal translation from the $(x-\frac{\pi }{2})$ term: A horizontal translation by $\frac{\pi }{2}$ units to the right.

4. Identify the vertical stretch from the multiplier $3$ outside the function: A vertical stretch by scale factor $3$.

5. Identify the vertical translation from the $+1$ at the end: A vertical translation by $1$ unit upwards.

6. State the full sequence. (Note: stretches should generally be applied before translations unless bracketed otherwise).