Transformations H G Ecd670

1. **State the problem:**\nWe have two functions:\n$$g(x) = (x+1)^2(x-2)^2$$\nand\n$$h(x) = (x+1)(x-1)(x+2)(x-2)$$\nWe want to find transformations of $h$ so that:\n\ni. Using translations only, $h$ has a local maximum at the same coordinates as $g$.\nii. Using dilation and translations, $h$ has both local minimums at the same coordinates as $g$.\n\n2. **Understand the functions:**\n- $g(x)$ is a quartic with double roots at $x=-1$ and $x=2$, so it has local extrema at these points.\n- $h(x)$ is a quartic with roots at $x=-2,-1,1,2$. It has local maxima and minima at different points than $g$.\n\n3. **Find local extrema of $g$:**\nSince $g(x) = (x+1)^2(x-2)^2$, its local extrema occur at $x=-1$ and $x=2$.\nCalculate $g(-1)$ and $g(2)$:\n$$g(-1) = 0, \quad g(2) = 0$$\nSo local extrema coordinates are $(-1,0)$ and $(2,0)$.\n\n4. **Find local extrema of $h$:**\n$h(x) = (x+1)(x-1)(x+2)(x-2) = (x^2-1)(x^2-4) = x^4 - 5x^2 + 4$\nDerivative:\n$$h'(x) = 4x^3 - 10x$$\nSet $h'(x)=0$:\n$$4x^3 - 10x = 0 \Rightarrow 2x(2x^2 - 5) = 0$$\nRoots:\n$$x=0, \quad x=\pm \sqrt{\frac{5}{2}}$$\nCalculate $h$ at these points:\n$$h(0) = 4$$\n$$h(\pm \sqrt{\frac{5}{2}}) = \left(\frac{5}{2}\right)^2 - 5 \cdot \frac{5}{2} + 4 = \frac{25}{4} - \frac{25}{2} + 4 = \frac{25}{4} - \frac{50}{4} + \frac{16}{4} = -\frac{9}{4} = -2.25$$\nSo local maximum at $(0,4)$ and local minima at $\left(\pm \sqrt{\frac{5}{2}}, -2.25\right)$.\n\n5. **Part i: Using translations only to align local maximum of $h$ with that of $g$:**\n- $h$ has local max at $(0,4)$, $g$ has local max at $(-1,0)$.\n- Translation moves points by $(x,y) \to (x', y') = (x + h_x, y + h_y)$.\n- To move $(0,4)$ to $(-1,0)$, set:\n$$x' = x - 1$$\n$$y' = y - 4$$\n- So the transformation is:\n$$(x,y) \to (x-1, y-4)$$\n- In function notation, new function $H_1(x) = h(x+1) - 4$.\n\n6. **Part ii: Using dilation and translations to align both local minima of $h$ with those of $g$:**\n- $h$ local minima at $\left(\pm \sqrt{\frac{5}{2}}, -2.25\right)$\n- $g$ local minima at $(-1,0)$ and $(2,0)$\n- We want to find $n, a, h, k$ such that:\n$$(x,y) \to (n x + h, a y + k)$$\n- Map $x$-coordinates of $h$ minima to $g$ minima:\n$$n \cdot \sqrt{\frac{5}{2}} + h = 2$$\n$$n \cdot \left(-\sqrt{\frac{5}{2}}\right) + h = -1$$\n- Subtract second from first:\n$$n \sqrt{\frac{5}{2}} - n \left(-\sqrt{\frac{5}{2}}\right) = 2 - (-1)$$\n$$2 n \sqrt{\frac{5}{2}} = 3 \Rightarrow n = \frac{3}{2 \sqrt{\frac{5}{2}}} = \frac{3}{2} \cdot \frac{1}{\sqrt{\frac{5}{2}}} = \frac{3}{2} \cdot \sqrt{\frac{2}{5}} = \frac{3}{2} \cdot \frac{\sqrt{2}}{\sqrt{5}} = \frac{3 \sqrt{2}}{2 \sqrt{5}}$$\n- Find $h$ by plugging back:\n$$n \sqrt{\frac{5}{2}} + h = 2 \Rightarrow h = 2 - n \sqrt{\frac{5}{2}}$$\nCalculate $n \sqrt{\frac{5}{2}}$:\n$$n \sqrt{\frac{5}{2}} = \frac{3 \sqrt{2}}{2 \sqrt{5}} \cdot \sqrt{\frac{5}{2}} = \frac{3 \sqrt{2}}{2 \sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{2}} = \frac{3 \sqrt{2} \sqrt{5}}{2 \sqrt{5} \sqrt{2}} = \frac{3}{2} = 1.5$$\nSo\n$$h = 2 - 1.5 = 0.5$$\n- Now map $y$-coordinates:\nWe want:\n$$a \cdot (-2.25) + k = 0$$\n- We have two points with same $y$-value $-2.25$ mapped to $0$, so this single equation suffices.\n- Choose $a=1$ for simplicity, then $k=2.25$.\n- So the transformation is:\n$$(x,y) \to \left(\frac{3 \sqrt{2}}{2 \sqrt{5}} x + 0.5, y + 2.25\right)$$\n- In function form, new function $H_2(x) = h\left(\frac{2 \sqrt{5}}{3 \sqrt{2}} (x - 0.5)\right) + 2.25$.\n\n7. **Summary of transformations in mapping notation:**\n- Part i (translations only):\n$$(x,y) \to (x-1, y-4)$$\n- Part ii (dilation and translations):\n$$(x,y) \to \left(\frac{3 \sqrt{2}}{2 \sqrt{5}} x + 0.5, y + 2.25\right)$$\n\n8. **Final answers:**\n- i. Translate $h$ by $-1$ in $x$ and $-4$ in $y$ to align local max with $g$.\n- ii. Dilate $x$ by $\frac{3 \sqrt{2}}{2 \sqrt{5}}$, translate $x$ by $0.5$, and translate $y$ by $2.25$ to align local minima with $g$.