lecture 5 transformations ♡

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

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

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

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

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

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

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

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.

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

$f (x) = (x - 2) (x + 4)$ . 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 $y = 3 x^{2} + 2$ after dilation of 3 from y axis?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Find the original range.

$f^{'} (x) = 3 x^{2} + 1$ , and we know that $x \geq 0,$ meaning $f^{'} (x) \geq 0$ . 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.

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

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

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

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

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

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

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

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

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 $y = 6 ln (x + 2) - 1$ .

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

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

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

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

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

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

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

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

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

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

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

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

$f^{'} (- 1) = 3$

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

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 $y = tan (5 x - 2) + 4$ onto $y = 2 tan (3 x)$ ?

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

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

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

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

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

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

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

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

$tan (\frac{1}{3} x - \frac{π}{6}) = tan (3 x^{'} - \frac{π}{3})$

$\frac{1}{3} x - \frac{π}{6} = 3 x^{'} - \frac{π}{3}$
$3 x^{'} = \frac{1}{3} x + \frac{π}{6}$
$x^{'} = \frac{1}{9} x + \frac{π}{18}$

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

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

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

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

Consider the transformation $T (x, y) = (3 + x, 4 - y)$ . Find the inverse transformation $T^{- 1}$ .

$x^{'} = 3 + x$
$y^{'} = 4 - y$

$x = x - 3$
$y = 4 - y^{'}$

Therefore.
$T^{- 1} (x, y) = (x - 3, 4 - y^{'})$

Consider the transformation $T (x, y) = (\frac{3 x + 4}{3}, \frac{5 y - 4}{4})$ . Find the inverse transformation $T^{- 1}$ .

$x^{'} = \frac{3 x + 4}{3}$
$y^{'} = \frac{5 y - 4}{4}$

$x = \frac{3 x - 4}{3}$
$y = \frac{4 y + 4}{5}$

Therefore.

$T^{- 1} (x, y) = (\frac{3 x - 4}{3}, \frac{4 y + 4}{5})$

Let $f (x) = x^{2} - 4$ and $g (x) = 4 (x - 1)^{2} - 4$ , let the graph of g be a transformation of the graph of f where the transformations have been applied in the following order: Dilation $n$ from the x axis, translation 1 right and $m$ up. $n$ and $m$ are positive real numbers. What is $m - n$ ?

Let f be y and g be y’

$y^{'} = 4 (x^{'} - 1)^{2} - 4$

$(x, y) \to x^{'} = x + 1, y^{'} = n y + m$
$x = x^{'} - 1$
$y = \frac{y ^{'} - m}{n}$

$y = x^{2} - 4$

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

$4 (x^{'} - 1)^{2} - 4 = n (x^{'} - 1)^{2} - 4 n + m$
$4 (x - 1)^{2} - 4 = n (x^{2} - 2 x + 1) - 4 n + m$
$4 (x - 1)^{2} - 4 = n x^{2} - 2 n x + n - 4 n + m$
$4 x^{2} - 8 x = n x^{2} - 2 n x - 3 n + m$

$4 = n$
$- 8 = - 2 n$
$- 3 n + m = 0$

$- 12 + m = 0$
$m = 12$
$m - n = 8$

Given that $g (x) = 2 x - 2$ and $f (x) = x^{2} + 2 x + 3$ , for what dilation factor(s) from the x-axis does the graph of g(x) need to undergo if g(x) and f(x) intersect exactly once?

Let dilation factor needed be d
Let y be g(x) and dilated g(x) be y’

$y = 2 x - 2$
$(x, y) \to x^{'} = x, y^{'} = d y$
$\frac{y ^{'}}{d} = 2 x^{'} - 2$

$y^{'} = 2 d x^{'} - 2 d$

Intersection.

$x^{2} + 2 x + 3 = 2 d x - 2 d$
$x^{2} + (2 - 2 d) x + 3 + 2 d = 0$

There is one solution to this equation. Discriminant be zero.
$Δ = 0$
$Δ = (2 - 2 d)^{2} - 4 (1) (3 + 2 d)$
$0 = 4 - 8 d + 4 d^{2} - 12 - 8 d$
$0 = 4 d^{2} - 16 d - 8$
$0 = d^{2} - 4 d - 2$

Use quadratic formula.

$d = \frac{4 \pm 16 - 4 ( - 2 )}{2}$
$d = 2 \pm 6$

The word dilation implies the dilation factor is positive, therefore.

$d = 2 + 6$

wiki ♡

Explorer

lecture 5 transformations

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

Consider the transformation $T (x, y) = (3 + x, 4 - y)$ . Find the inverse transformation $T^{- 1}$ .

Consider the transformation $T (x, y) = (\frac{3 x + 4}{3}, \frac{5 y - 4}{4})$ . Find the inverse transformation $T^{- 1}$ .

Given that $g (x) = 2 x - 2$ and $f (x) = x^{2} + 2 x + 3$ , for what dilation factor(s) from the x-axis does the graph of g(x) need to undergo if g(x) and f(x) intersect exactly once?

Table of Contents

Graph View

Table of Contents

wiki ♡

Explorer

lecture 5 transformations

What is y=2x2 after dilation 2 in x axis?

y=(2x−1)2−3​+5. Translate 1 right and 7 down. Then dilate 21​ from x axis and dilate 2 from y axis. Then reflect in y axis.

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

The point (9,11) lies on f. The graph is dilated by 31​ from the x axis and translated down 5 units. Find the image of the point.

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?

What is y=3x2+2 after dilation of 3 from y axis?

What is y=3x−9​ after reflection in x axis, then dilation 31​ in y axis.

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

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

What is y=−1−x​+1 after dilation 5 from y axis, then reflection in y axis, then translate 2 left and 8 up.

What is y=x4+6x2+9 after translation 1 up, then reflection in y axis, then dilation 2 from x axis.

f:[0,∞)→R,f(x)=x3+x. What is the range of the image function after reflection in x axis?

The point (−1,21​) lies on d. The graph is reflected in both axes , then translated 1 left. Find the image of the point.

Describe the transformations that map y=(2x+1)24​−6 onto y=(x−1)210​+3.

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)?

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

What is y=6x2 after dilation 2 in y axis?

The function f(x) is transformed by T(x,y) into −a2f(bx−c2)−d. Describe the transformations.

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=−221​f(−x+2)+1.

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

Describe the transformations that map y=6tan(31​x−6π​)+5 onto y=2tan(3x−3π​)+7.

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

Consider the transformation T(x,y)=(3+x,4−y). Find the inverse transformation T−1.

Consider the transformation T(x,y)=(33x+4​,45y−4​). Find the inverse transformation T−1.

Let f(x)=x2−4 and g(x)=4(x−1)2−4, let the graph of g be a transformation of the graph of f where the transformations have been applied in the following order: Dilation n from the x axis, translation 1 right and m up. n and m are positive real numbers. What is m−n?

Given that g(x)=2x−2 and f(x)=x2+2x+3, for what dilation factor(s) from the x-axis does the graph of g(x) need to undergo if g(x) and f(x) intersect exactly once?

Table of Contents

Graph View

Table of Contents

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

Consider the transformation $T (x, y) = (3 + x, 4 - y)$ . Find the inverse transformation $T^{- 1}$ .

Consider the transformation $T (x, y) = (\frac{3 x + 4}{3}, \frac{5 y - 4}{4})$ . Find the inverse transformation $T^{- 1}$ .

Given that $g (x) = 2 x - 2$ and $f (x) = x^{2} + 2 x + 3$ , for what dilation factor(s) from the x-axis does the graph of g(x) need to undergo if g(x) and f(x) intersect exactly once?