Functions: Question 4

Worked solution

Part (a): $y = f(x) + 6$

Adding a constant outside the function translates the graph vertically: $y=f(x)+a$ shifts every point up by $a$ (for $a>0$), leaving the $x$-coordinate unchanged.

Here $a=6$, so the minimum point moves from $(4,-3)$ to: $(4,\ -3+6) = (4, 3)$

Part (b): $y = f(x-2)$

Adding a constant inside the function translates the graph horizontally: $y=f(x-c)$ shifts every point $c$ units to the right (for $c>0$), leaving the $y$-coordinate unchanged.

Here $c=2$, so the minimum point moves from $(4,-3)$ to: $(4+2,\ -3) = (6, -3)$

Part (c): $y = 2f(x) - 1$

This combines two transformations applied to the output of $f$: a vertical stretch of scale factor $2$, then a vertical translation of $-1$. Multiplying by a positive constant and then subtracting a constant doesn’t change which $x$-value gives the smallest output, so the minimum still occurs at $x=4$.

Take the original minimum value $f(4)=-3$ and apply the same transformation to it:

$y = 2(-3) - 1 = -6-1 = -7$

So the minimum point is: $(4, -7)$

Part (d): $y = f(3x)$

Replacing $x$ with $3x$ inside the function is a horizontal stretch of scale factor $\dfrac{1}{3}$. Every point moves $3$ times closer to the $y$-axis. The $y$-coordinate of the minimum is unchanged, but the $x$-coordinate at which it occurs is divided by $3$.

The original minimum occurs at $x=4$. For $y=f(3x)$ to reach that same minimum, we need $3x=4$, i.e. $x = \dfrac{4}{3}$, and the $y$-value there is still $f(4)=-3$.

So the minimum point is: $\left(\frac{4}{3},\ -3\right)$

Final answers

• (a) $(4, 3)$
• (b) $(6, -3)$
• (c) $(4, -7)$
• (d) $\left(\boxed{\dfrac{4}{3}},\ -3\right)$