lecture 5 transformations

What is $y=2x^2$ after dilation 2 in x axis?

$x,y\to x^{\prime }=x,y^{\prime }=2y$
$x=x^{\prime },y=\frac{1}{2}{y}^{\prime }$
$y=2x^2$
$\to$
$\frac{1}{2}y=2x^2$
$y=4x^2$

$y=\frac{-3}{(2x-1{)}^2}+5$. Translate 1 right and 7 down. Then dilate $\frac{1}{2}$ from x axis and dilate 2 from y axis. Then reflect in y axis.

$x^{\prime }=x,y^{\prime }=y$
$\to$
$x^{\prime }=x+1,y^{\prime }=y-7$
$\to$
$x^{\prime }=2(x+1),y^{\prime }=\frac{1}{2}(y-7)$
$\to$
$x^{\prime }=-2(x+1),y^{\prime }=\frac{1}{2}(y-7)$
$x=\frac{x^{\prime }}{-2}-1$
$y=2y^{\prime }+7$

$y=\frac{-3}{(2x-1{)}^2}+5$
$\to$
$2y+7=\frac{-3}{(-x-3{)}^2}+5$
$2y+7=\frac{-3}{(-1(x+3){)}^2}+5$
$2y+7=\frac{-3}{(x+3{)}^2}+5$
$2y=\frac{-3}{(x+3{)}^2}-2$
$y=\frac{-3}{2(x+3{)}^2}-1$

The graph of $y=2x$ is translated 1 right and 3 up, what is the resulting function?

$x,y\to x^{\prime }=x+1,y^{\prime }=y+3$
$x=x^{\prime }-1,y=y^{\prime }-3$
$y=2x$
$y^{\prime }-3=2x^{\prime }-2$
$\to$
$y=2x+1$

The point $(9,11)$ lies on f. The graph is dilated by $\frac{1}{3}$ from the x axis and translated down 5 units. Find the image of the point.

$x,y\to x^{\prime }=x,y^{\prime }=\frac{1}{3}y-5$
$(9,11)\to (9,\frac{-4}{3})$

$f(x)=(x-2)(x+4)$. If a horizontal translation in the form $y=f(x-k)$ is applied, what values of k will the function have 2 positive solutions?

The original function has roots at x=2 and x=-4. We will need a horizontal translation of more than 4 right.

$k>4$

What is $y=3x^2+2$ after dilation of 3 from y axis?

$x,y\to x^{\prime }=3x,y^{\prime }=y$
$x=\frac{x^{\prime }}{3}$
$y=3x^2+2$
$\to$
$y=\frac{x^2}{3}+2$

What is $y=\sqrt{3x-9}$ after reflection in x axis, then dilation $\frac{1}{3}$ in y axis.

$x,y\to x^{\prime }=\frac{1}{3}x,y^{\prime }=-y$
$x=3x^{\prime },y=-y^{\prime }$
$y=\sqrt{3x-9}$
$\to$
$-y=\sqrt{3(3x)-9}$
$y=-\sqrt{9x-9}$

The transformation $T(x,y)=(-2x+6,3y-2)$ transforms a graph into $y=4x-12$. What is the pre-image?

$x,y\to x^{\prime }=-2x+6,y^{\prime }=3y-2$

$y^{\prime }=4x^{\prime }-12$
$\to$
$3y-2=4(-2x+6)-12$
$3y-2=-8x+24-12$
$3y=-8x+14$
$y=\frac{-8}{3}+\frac{14}{3}$

The transformation $T(x,y)=(3x+2,-y+4)$ transforms a graph into $y=\frac{(x-3{)}^2}{6}+2$. What is the pre-image?

$x,y\to x^{\prime }=3x+2,y^{\prime }=-y+4$

$y^{\prime }=\frac{(x^{\prime }-3{)}^2}{6}+2$
$\to$
$-y+4=\frac{(3x+2-3{)}^2}{6}+2$
$-y=\frac{(3x-1{)}^2}{6}-2$
$y=\frac{-(3x-1{)}^2}{6}+2$

What is $y=-\sqrt{1-x}+1$ after dilation 5 from y axis, then reflection in y axis, then translate 2 left and 8 up.

$x,y\to x^{\prime }=-5x-2,y^{\prime }=y+8$
$x=\frac{x^{\prime }+2}{-5}$
$y=y^{\prime }-8$

$y=-\sqrt{1-x}+1$
$\to$
$y-8=-\sqrt{1-\frac{x+2}{-5}}+1$
$y-8=-\sqrt{\frac{5}{5}+\frac{x+2}{5}}+1$
$y=-\sqrt{\frac{x+7}{5}}+9$

What is $y=x^4+6x^2+9$ after translation 1 up, then reflection in y axis, then dilation 2 from x axis.

$x^{\prime }=x,y^{\prime }=y$
$\to$
$x^{\prime }=x+1,y^{\prime }=y$
$\to$
$x^{\prime }=-(x+1),y^{\prime }=y$
$\to$
$x^{\prime }=-(x+1),y^{\prime }=2y$

$x=-x^{\prime }-1$
$y=\frac{y^{\prime }}{2}$

$y=x^4+6x^2+9$
$\to$
$\frac{y}{2}=(-x-1{)}^4+6(-x-1{)}^2+9$
$y=2(x+1{)}^4+12(x+1{)}^2+18$
$y=2(x^4+4x^3+4x^2+1)+12(x^2+2x+1)+18$
$y=2x^4+8x^3+24x^2+32x+32$

$f:[0,\infty )\to \mathbb{R},f(x)=x^3+x$. What is the range of the image function after reflection in x axis?

The original domain is $[0,\infty ]$.

Find the original range.

$f^{\prime }(x)=3x^2+1$, and we know that $x\ge 0,$ meaning $f^{\prime }(x)\ge 0$. The graph is strictly increasing.

$f(0)=0$

So the range is $[0,\infty )$.

After the reflection, the range is $(-\infty ,0]$.

The point $(-1,\frac{1}{2})$ lies on d. The graph is reflected in both axes , then translated 1 left. Find the image of the point.

$x,y\to x^{\prime }=-x-1,y^{\prime }=-y$

$(-1,\frac{1}{2})\to (0,-\frac{1}{2})$

Describe the transformations that map $y=\frac{4}{(2x+1{)}^2}-6$ onto $y=\frac{10}{(x-1{)}^2}+3$.

$y^{\prime }=\frac{10}{(x^{\prime }-1{)}^2}+3$
$\frac{y^{\prime }-3}{10}=\frac{1}{(x^{\prime }-1{)}^2}$

$y=\frac{4}{(2x+1{)}^2}-6$

$y+6=\frac{4}{(2x+1{)}^2}$
$\frac{y+6}{4}=\frac{1}{(2x+1{)}^2}$
$\to$
$\frac{y^{\prime }-3}{10}=\frac{1}{(x^{\prime }-1{)}^2}$

$\frac{y+6}{4}=\frac{y^{\prime }-3}{10}$
$10y+60=4y^{\prime }-12$
$y^{\prime }=\frac{10y+72}{4}$
$y^{\prime }=\frac{5}{2}y+18$

$\frac{1}{(2x+1{)}^2}=\frac{1}{(x^{\prime }-1{)}^2}$
$2x+1=x^{\prime }-1$
$x^{\prime }=2x+2$

• Dilate 2 from y axis.
• Dilate $\frac{5}{2}$ from x axis.
• Translate 2 right and 18 up.

$f(x)=(x-3{)}^3+3$. If vertical translation of $k$ is applied, what values of $k$ will the x-intercept be at $(3,0)$?

We want to solve the equation $0=(3-3{)}^3+3+k$.
$k=-3$

Describe the transformations that map $y=3\ln (7-x)+2$ onto $y=6\ln (x+2)-1$.

$y^{\prime }=6\ln (x^{\prime }+2)-1$
$\frac{y^{\prime }+1}{6}=\ln (x^{\prime }+2)$

$y=3\ln (7-x)+2$
$\frac{y-2}{3}=\ln (7-x)$
$\to$
$\frac{y^{\prime }+1}{6}=\ln (x^{\prime }+2)$

$\frac{y-2}{3}=\frac{y^{\prime }+1}{6}$
$2y-4=y^{\prime }+1$
$y^{\prime }=2y-5$

$\ln (7-x)=\ln (x^{\prime }+2)$
$7-x=x^{\prime }+2$
$x^{\prime }=-x+5$

• Dilate 2 from x axis.
• Reflect in y axis.
• Translate 5 right and 5 down.

What is $y=6x^2$ after dilation 2 in y axis?

$x,y\to x^{\prime }=2x,y^{\prime }=y$
$x=\frac{x^{\prime }}{2}$
$y=6x^2$
$\to$
$y=6(\frac{x}{2}{)}^2$
$y=\frac{3x^2}{2}$

The function $f(x)$ is transformed by $T(x,y)$ into $-a^{2}f(bx-c^2)-d$. Describe the transformations.

We can see that $x\to x^{\prime }=bx-c^2$.
And $f\to f^{\prime }=-a^{2}f-d$.

The tangent at the point $(-1,0)$ on the curve $y=f(x)$ has the equation $y=3x+c$. Find the equation of the tangent at the point $(3,1)$ to the curve $y=-2\frac{1}{2}f(-x+2)+1$.

Let the transformed function be $g(x)=-2\frac{1}{2}f(-x+2)+1$

$g^{\prime }(x)=-\frac{1}{2}\cdot f^{\prime }(-x+2)\cdot -1$
$g^{\prime }(x)=\frac{1}{2}{f}^{\prime }(-x+2)$

$g^{\prime }(3)=\frac{1}{2}{f}^{\prime }(-3+2)$
$g^{\prime }(3)=\frac{1}{2}{f}^{\prime }(-1)$

$f^{\prime }(-1)=3$

$g^{\prime }(3)=\frac{3}{2}$

Now find the line. Use the point $(3,1)$ and $m=\frac{3}{2}$

$y-y_1=m(x-x_1)$

$y-1=\frac{3}{2}(x-3)$
$y=\frac{3}{2}x-\frac{9}{2}+1$
$y=\frac{3}{2}x-\frac{7}{2}$

What is the transformation $T$ that maps $y=\tan (5x-2)+4$ onto $y=2\tan (3x)$?

$y^{\prime }=2\tan (3x^{\prime })$
$\frac{y^{\prime }}{2}=\tan (3x^{\prime })$

$y=\tan (5x-2)+4$
$y-4=\tan (5x-2)$
$\to$
$\frac{y^{\prime }}{2}=\tan (3x^{\prime })$

$\frac{y^{\prime }}{2}=y-4$
$y^{\prime }=2y-8$
$5x-2=3x^{\prime }$
$x^{\prime }=\frac{5}{3}x-\frac{2}{3}$

$T(x,y)=(\frac{5}{3}x-\frac{2}{3},2y-8)$

Describe the transformations that map $y=6\tan (\frac{1}{3}x-\frac{\pi }{6})+5$ onto $y=2\tan (3x-\frac{\pi }{3})+7$.

$y^{\prime }=2\tan (3x^{\prime }-\frac{\pi }{3})+7$
$\frac{y^{\prime }-7}{2}=\tan (3x^{\prime }-\frac{\pi }{3})$

$y=6\tan (\frac{1}{3}x-\frac{\pi }{6})+5$
$\frac{y-5}{6}=\tan (\frac{1}{3}x-\frac{\pi }{6})$
$\to$
$\frac{y^{\prime }-7}{2}=\tan (3x^{\prime }-\frac{\pi }{3})$

$\frac{y^{\prime }-7}{2}=\frac{y-5}{6}$
$3y^{\prime }-21=y-5$
$y^{\prime }=\frac{y}{3}+\frac{16}{3}$

$\tan (\frac{1}{3}x-\frac{\pi }{6})=\tan (3x^{\prime }-\frac{\pi }{3})$

$\frac{1}{3}x-\frac{\pi }{6}=3x^{\prime }-\frac{\pi }{3}$
$3x^{\prime }=\frac{1}{3}x+\frac{\pi }{6}$
$x^{\prime }=\frac{1}{9}x+\frac{\pi }{18}$

• Dilate $\frac{1}{9}$ from y axis.
• Dilate $\frac{1}{3}$ from x axis.
• Translate $\frac{\pi }{18}$ right and $\frac{16}{3}$ up.

What is $y=(x+4{)}^2+2$ after dilation 3 from x axis?

$x,y\to x^{\prime }=x,y^{\prime }=3y$
$y=\frac{y^{\prime }}{3}$

$y=(x+4{)}^2+2$
$\to$
$\frac{y}{3}=(x+4{)}^2+2$
$y=3(x+4{)}^2+6$