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Domain:
(-oo,2)uu(2,oo)

x and
y-intercepts:
(0,0)
Vertical asymptote:
x=2
Horizontal asymptote:
y=1

f decreasing:
(-oo,0) and
(2,oo)

f increasing:
(0,2)
Local minimum value:
f(0)=0
No local maximum values
Concavity down:
(-oo,-1)
Concavity up:
(-1,2) and
(2,oo)
Inflection points:
(-1,(1)/(9))

- Domain: $(-\infty, 2) \cup(2, \infty)$ $\newline$ - $x$ and $y$ -intercepts: $(0,0)$ $\newline$ - Vertical asymptote: $x=2$ $\newline$ - Horizontal asymptote: $y=1$ $\newline$ - $f$ decreasing: $(-\infty, 0)$ and $(2, \infty)$ $\newline$ - $f$ increasing: $x$ $0$ $\newline$ - Local minimum value: $x$ $1$ $\newline$ - No local maximum values $\newline$ - Concavity down: $x$ $2$ $\newline$ - Concavity up: $x$ $3$ and $(2, \infty)$ $\newline$ - Inflection points: $x$ $5$

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Function transformation rules

Full solution

Q. - Domain:

(-\infty, 2) \cup(2, \infty)

\newline

-

x

and

y

-intercepts:

(0,0)

\newline

- Vertical asymptote:

x=2

\newline

- Horizontal asymptote:

y=1

\newline

-

f

decreasing:

(-\infty, 0)

and

(2, \infty)

\newline

-

f

increasing:

x

0

\newline

- Local minimum value:

x

1

\newline

- No local maximum values

\newline

- Concavity down:

x

2

\newline

- Concavity up:

x

3

and

(2, \infty)

\newline

- Inflection points:

x

5

Start with Rational Function Form: Given the domain, $x$ -intercepts, and asymptotes, we can start by considering the rational function form $f(x) = \frac{a}{(x-h)} + k$ , where ' $a$ ' affects the vertical stretch, ' $h$ ' is the horizontal shift (related to vertical asymptote), and ' $k$ ' is the vertical shift (related to horizontal asymptote).
Vertical Asymptote Determination: Since the vertical asymptote is $x=2$ , we know $h=2$ . This gives us a denominator of $(x-2)$ .
Horizontal Asymptote Determination: The horizontal asymptote is $y=1$ , which means $k=1$ . So, the function so far looks like $f(x) = \frac{a}{(x-2)} + 1$ .
X-Intercept Calculation: The x-intercept is at $(0,0)$ , which means the function must equal zero when $x=0$ . Plugging in, we get $0 = \frac{a}{(0-2)} + 1$ . Solving for ' $a$ ' gives us $a = -2$ .
Function Adjustment: Now we have the function $f(x) = -\frac{2}{(x-2)} + 1$ . We need to check if this function matches the other given properties.
Decreasing Function Check: Check if $f$ is decreasing on $(-\infty,0)$ and $(2,\infty)$ . Since $'a'$ is negative, the function is decreasing where the denominator is positive, which is when $x > 2$ . It's also decreasing when $x < 0$ because the negative $'a'$ flips the positive slope of $1/(x-2)$ to negative. So this checks out.
Increasing Function Check: Check if $f$ is increasing on $(0,2)$ . Since $a$ is negative, the function is increasing where the denominator is negative, which is when $0 < x < 2$ . This checks out.
Local Minimum Verification: Check the local minimum value. Since the function is decreasing up to $x=0$ and increasing after, $f(0)=0$ is indeed a local minimum. This checks out.
Concavity Analysis: Check concavity. The second derivative of $f(x)$ would show concavity. Since we're not explicitly finding the second derivative, we'll rely on the given information. The function is concave down on $(-\infty,-1)$ and concave up on $(-1,2)$ and $(2,\infty)$ . This suggests an inflection point at $x=-1$ .
Inflection Point Calculation: Find the $y$ -coordinate of the inflection point by plugging $x=-1$ into $f(x)$ . $f(-1) = -\frac{2}{(-1-2)} + 1 = -\frac{2}{(-3)} + 1 = \frac{2}{3} + 1 = \frac{5}{3}$ . But the given inflection point is $(-1,\frac{1}{9})$ . This is a contradiction.

More problems from Function transformation rules

Question

g(x)

g(x)

is the reflection across the x-axis of

f(x)=9|x+2|-4

\newline

\newline

\text{}(A)g(x) = 9|x+2| - 4\text{]}

\newline

\text{}(B)g(x) = 9|x-2| - 4\text{]}

\newline

\text{}(C)g(x)=-9|x-2| +4\text{]}

\newline

\text{(D)}g(x) = -9|x + 2| + 4\text{]}

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What does the transformation

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\newline

\text{translates it } 4 \text{ units right}

\newline

\text{translates it } 4 \text{ units down}

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\text{translates it } 4 \text{ units up}

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\newline

[A] stretches it vertically

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[B] stretches it horizontally

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[C]shrinks it horizontally

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\newline

\newline

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\newline

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\newline

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\newline

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Question

What kind of transformation converts the graph of

f(x) = -3x^2 - 4

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\newline

\newline

\text{[A]translation 10 units up}

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\text{[B]translation 10 units down}

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\text{[D]translation 10 units right}

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Question

y=x^{2}

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y=x^{2}

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units and to the right by

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y=x^{2}

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y=x^{2}

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What is the equation of the new parabola?

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Question

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