CALCU-A family of curves is a set of curves whose

Generated Graph

Solution by Steps

step 1

The function given is

f(x) = x^2

step 2

To find the equation of the tangent line at

(3,9)

, we first need to find the derivative of

f(x)

step 3

The derivative of

f(x) = x^2

is

f'(x) = 2x

step 4

Evaluate the derivative at

x = 3

:

f'(3) = 2 \cdot 3 = 6

step 5

The slope of the tangent line at

(3,9)

is

6

step 6

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3, 9)

and

m = 6

:

step 7

y - 9 = 6(x - 3)

step 8

Simplify to get the equation of the tangent line:

y = 6x - 9

Answer

The equation of the tangent line to

f(x) = x^2

at

(3,9)

is

y = 6x - 9

.

(b) Explore the family of curves

g(x)=x^{2}+k

where

k \in R

# (i) Describe the effect of

k

on the graph of

g(x)=x^{2}

step 1

The parameter

k

in the function

g(x) = x^2 + k

represents a vertical shift of the graph of

f(x) = x^2

step 2

If k > 0, the graph shifts upward by

k

units. If k < 0, the graph shifts downward by

k

units

# (ii) Choose 4 different values of

k

to explore

g(x)=x^{2}+k

and sketch the graphs

step 1

Choose

k = 1, 2, 5, 8

step 2

The graphs of

g(x) = x^2 + 1

,

g(x) = x^2 + 2

,

g(x) = x^2 + 5

, and

g(x) = x^2 + 8

will be parabolas that are vertically shifted versions of

f(x) = x^2

# (iii) Find the tangent line equation to each curve in part ii. at

x=3

step 1

For

g(x) = x^2 + k

, the derivative is

g'(x) = 2x

step 2

Evaluate the derivative at

x = 3

:

g'(3) = 2 \cdot 3 = 6

step 3

The slope of the tangent line at

x = 3

is

6

for all values of

k

step 4

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3, 3^2 + k) = (3, 9 + k)

and

m = 6

:

step 5

y - (9 + k) = 6(x - 3)

step 6

Simplify to get the equation of the tangent line:

y = 6x - 9 + k

Answer

The tangent line equations to the curves

g(x) = x^2 + k

at

x = 3

are

y = 6x - 9 + k

.

(c) Repeat the exploration for the family of curves

q(x)=(x-h)^{2}

where

h \in R

step 1

The function given is

q(x) = (x - h)^2

step 2

To find the equation of the tangent line at

x = 3 + h

, we first need to find the derivative of

q(x)

step 3

The derivative of

q(x) = (x - h)^2

is

q'(x) = 2(x - h)

step 4

Evaluate the derivative at

x = 3 + h

:

q'(3 + h) = 2((3 + h) - h) = 2 \cdot 3 = 6

step 5

The slope of the tangent line at

x = 3 + h

is

6

step 6

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3 + h, (3 + h - h)^2) = (3 + h, 9)

and

m = 6

:

step 7

y - 9 = 6(x - (3 + h))

step 8

Simplify to get the equation of the tangent line:

y = 6x - 18 - 6h + 9 = 6x - 6h - 9

Answer

The tangent line equations to the curves

q(x) = (x - h)^2

at

x = 3 + h

are

y = 6x - 6h - 9

.

Key Concept

Tangent Line Equation

Explanation

The tangent line to a curve at a given point can be found using the derivative to determine the slope and the point-slope form of the line equation.

Solution by Steps

step 1

Given the family of curves

r(x) = a x^2

, we need to find the tangent line equation at

x = 3

step 2

First, find the derivative of

r(x)

with respect to

x

:

\frac{dr}{dx} = 2ax

step 3

Evaluate the derivative at

x = 3

:

\frac{dr}{dx} \bigg|_{x=3} = 2a(3) = 6a

step 4

The equation of the tangent line at

x = 3

is given by

y = r(3) + \frac{dr}{dx} \bigg|_{x=3} (x - 3)

step 5

Calculate

r(3)

:

r(3) = a(3)^2 = 9a

step 6

Substitute

r(3)

and

\frac{dr}{dx} \bigg|_{x=3}

into the tangent line equation:

y = 9a + 6a(x - 3)

step 7

Simplify the tangent line equation:

y = 9a + 6ax - 18a = 6ax - 9a

Answer

The tangent line equation at

x = 3

is

y = 6ax - 9a

.

Key Concept

Finding the tangent line equation

Explanation

To find the tangent line equation at a specific point, we need to evaluate the function and its derivative at that point and then use the point-slope form of the line equation.

Solution by Steps

step 1

Choose the function

y = \frac{1}{x}

and investigate the effect of transformations on its tangent line at

x = 1

step 2

Find the derivative of

y = \frac{1}{x}

:

\frac{dy}{dx} = -\frac{1}{x^2}

step 3

Evaluate the derivative at

x = 1

:

\frac{dy}{dx} \bigg|_{x=1} = -1

step 4

The equation of the tangent line at

x = 1

is given by

y = f(1) + \frac{dy}{dx} \bigg|_{x=1} (x - 1)

step 5

Calculate

f(1)

:

f(1) = \frac{1}{1} = 1

step 6

Substitute

f(1)

and

\frac{dy}{dx} \bigg|_{x=1}

into the tangent line equation:

y = 1 - 1(x - 1)

step 7

Simplify the tangent line equation:

y = 1 - x + 1 = 2 - x

Answer

The tangent line equation at

x = 1

for

y = \frac{1}{x}

is

y = 2 - x

.

Key Concept

Effect of transformations on tangent lines

Explanation

Investigating the effect of transformations involves finding the derivative of the function, evaluating it at the point of interest, and using the point-slope form to determine the tangent line equation.

how did u get The tangent line equation at x = 3 x=3 is y = 6 a x − 9 a y=6ax−9a.

Solution by Steps

step 1

To find the tangent line to the curve at

x = 3

, we first need to determine the derivative of the curve's equation, which represents the slope of the tangent line at any point

x

step 2

Let's assume the curve is given by

y = f(x)

. The derivative of

y

with respect to

x

is

f'(x)

step 3

Evaluate the derivative at

x = 3

to find the slope of the tangent line at this point. Suppose

f'(3) = m

step 4

The equation of the tangent line at

x = 3

can be written in the point-slope form:

y - y_1 = m(x - x_1)

, where

(x_1, y_1)

is the point of tangency

step 5

Given that the point of tangency is

(3, f(3))

, substitute

x_1 = 3

,

y_1 = f(3)

, and

m = f'(3)

into the point-slope form:

y - f(3) = f'(3)(x - 3)

step 6

Simplify the equation to get the tangent line in slope-intercept form:

y = f'(3)x - 3f'(3) + f(3)

step 7

According to the given information, the tangent line equation at

x = 3

is

y = 6ax - 9a

. This implies that

f'(3) = 6a

and

f(3) = -9a

Answer

The tangent line equation at

x = 3

is

y = 6ax - 9a

.

Key Concept

Derivative and Tangent Line Equation

Explanation

The derivative of a function at a point gives the slope of the tangent line at that point. Using the point-slope form of the line equation, we can derive the tangent line equation.

Generated Graph

Solution by Steps

step 1

To find the equation of the tangent line to

f(x) = x^2

at

(3, 9)

, we first need to find the derivative of

f(x)

step 2

The derivative of

f(x) = x^2

is

f'(x) = 2x

step 3

Evaluate the derivative at

x = 3

:

f'(3) = 2 \cdot 3 = 6

. This is the slope of the tangent line

step 4

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3, 9)

and

m = 6

, we get:

y - 9 = 6(x - 3)

step 5

Simplify the equation:

y - 9 = 6x - 18 \implies y = 6x - 9

Answer

The equation of the tangent line to

f(x) = x^2

at

(3, 9)

is

y = 6x - 9

.

Key Concept

Derivative and Tangent Line

Explanation

The derivative of a function at a point gives the slope of the tangent line at that point. Using the point-slope form of a line, we can find the equation of the tangent line.

---

step 1

To explore the family of curves

g(x) = x^2 + k

where

k \in \mathbb{R}

, we first describe the effect of

k

on the graph of

g(x) = x^2

step 2

The parameter

k

vertically shifts the graph of

x^2

up or down. For k > 0, the graph shifts up; for k < 0, the graph shifts down

step 3

Choose 4 different values of

k

(e.g.,

k = 1, 2, 5, 8

) and sketch the graphs on the same set of axes

step 4

For each curve

g(x) = x^2 + k

, find the tangent line equation at

x = 3

step 5

The derivative of

g(x) = x^2 + k

is

g'(x) = 2x

step 6

Evaluate the derivative at

x = 3

:

g'(3) = 2 \cdot 3 = 6

. This is the slope of the tangent line for all

k

step 7

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3, 3^2 + k) = (3, 9 + k)

and

m = 6

, we get:

y - (9 + k) = 6(x - 3)

step 8

Simplify the equation:

y - 9 - k = 6x - 18 \implies y = 6x - 9 - k

Answer

The tangent line equations for

g(x) = x^2 + k

at

x = 3

are

y = 6x - 9 - k

.

Key Concept

Effect of Vertical Shifts on Tangent Lines

Explanation

The parameter

k

shifts the graph vertically, affecting the y-intercept of the tangent line but not its slope.

---

step 1

To explore the family of curves

q(x) = (x - h)^2

where

h \in \mathbb{R}

, we find the tangent line equations at

x = 3 + h

step 2

The derivative of

q(x) = (x - h)^2

is

q'(x) = 2(x - h)

step 3

Evaluate the derivative at

x = 3 + h

:

q'(3 + h) = 2((3 + h) - h) = 2 \cdot 3 = 6

. This is the slope of the tangent line for all

h

step 4

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3 + h, (3 + h - h)^2) = (3 + h, 9)

and

m = 6

, we get:

y - 9 = 6(x - (3 + h))

step 5

Simplify the equation:

y - 9 = 6x - 18 - 6h \implies y = 6x - 9 - 6h

Answer

The tangent line equations for

q(x) = (x - h)^2

at

x = 3 + h

are

y = 6x - 9 - 6h

.

Key Concept

Effect of Horizontal Shifts on Tangent Lines

Explanation

The parameter

h

shifts the graph horizontally, affecting the x-intercept of the tangent line but not its slope.

---

step 1

To explore the family of curves

r(x) = ax^2

where

a \in \mathbb{R}

, we find the tangent line equations at

x = 3

step 2

The derivative of

r(x) = ax^2

is

r'(x) = 2ax

step 3

Evaluate the derivative at

x = 3

:

r'(3) = 2a \cdot 3 = 6a

. This is the slope of the tangent line

step 4

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (3, a \cdot 3^2) = (3, 9a)

and

m = 6a

, we get:

y - 9a = 6a(x - 3)

step 5

Simplify the equation:

y - 9a = 6ax - 18a \implies y = 6ax - 9a

Answer

The tangent line equations for

r(x) = ax^2

at

x = 3

are

y = 6ax - 9a

.

Key Concept

Effect of Scaling on Tangent Lines

Explanation

The parameter

a

scales the graph vertically, affecting both the slope and the y-intercept of the tangent line.

---

step 1

To investigate the effect of various transformations on the graph

y = \frac{1}{x}

and its corresponding tangent lines using

x = 1

as the original point of contact, we first find the tangent line equation to

y = \frac{1}{x}

at

x = 1

step 2

The derivative of

y = \frac{1}{x}

is

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

step 3

Evaluate the derivative at

x = 1

:

y'(1) = -1

. This is the slope of the tangent line

step 4

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1)

, where

(x_1, y_1) = (1, 1)

and

m = -1

, we get:

y - 1 = -1(x - 1)

step 5

Simplify the equation:

y - 1 = -x + 1 \implies y = -x + 2

Answer

The tangent line equation to

y = \frac{1}{x}

at

x = 1

is

y = -x + 2

.

Key Concept

Effect of Transformations on Rational Functions

Explanation

Transformations of rational functions affect the slope and intercepts of their tangent lines, which can be analyzed using derivatives.

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