How to sketch a graph of the rational function $r(x) = frac{x^2+6x-8x}{2x^2+10x+12}$
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How do I sketch a graph of the rational function: $r(x) = frac{x^2+6x-8x}{2x^2+10x+12}$ and how do I find the x-intercepts, Y intercept, vertical asympyotes, and horizontal asympyotes.
I know the basic definition of the terms, but this is my first time doing a problem like this and I really want to see if anyone could break it down.
algebra-precalculus
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up vote
-1
down vote
favorite
How do I sketch a graph of the rational function: $r(x) = frac{x^2+6x-8x}{2x^2+10x+12}$ and how do I find the x-intercepts, Y intercept, vertical asympyotes, and horizontal asympyotes.
I know the basic definition of the terms, but this is my first time doing a problem like this and I really want to see if anyone could break it down.
algebra-precalculus
Do have any guidelines? Sketching without a computer involves using derivatives (first and second).
– Vasya
Nov 23 at 19:44
1
Denominator is $2(x+3)(x+2)$. This at least tells you the function is undefined at $x=-3, x=-2$. (We want to know the domain of the function).
– amWhy
Nov 23 at 19:48
3
$x^2 + 6x - 8x = x^2 - 2x$. Did you mean $x^2 + 6x - 8$?
– N. F. Taussig
Nov 23 at 19:50
you can use this online graphing software to graph your function. desmos.com/calculator
– user25406
Nov 23 at 19:52
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How do I sketch a graph of the rational function: $r(x) = frac{x^2+6x-8x}{2x^2+10x+12}$ and how do I find the x-intercepts, Y intercept, vertical asympyotes, and horizontal asympyotes.
I know the basic definition of the terms, but this is my first time doing a problem like this and I really want to see if anyone could break it down.
algebra-precalculus
How do I sketch a graph of the rational function: $r(x) = frac{x^2+6x-8x}{2x^2+10x+12}$ and how do I find the x-intercepts, Y intercept, vertical asympyotes, and horizontal asympyotes.
I know the basic definition of the terms, but this is my first time doing a problem like this and I really want to see if anyone could break it down.
algebra-precalculus
algebra-precalculus
edited Nov 23 at 19:30
Key Flex
7,44941232
7,44941232
asked Nov 23 at 19:28
User231
165
165
Do have any guidelines? Sketching without a computer involves using derivatives (first and second).
– Vasya
Nov 23 at 19:44
1
Denominator is $2(x+3)(x+2)$. This at least tells you the function is undefined at $x=-3, x=-2$. (We want to know the domain of the function).
– amWhy
Nov 23 at 19:48
3
$x^2 + 6x - 8x = x^2 - 2x$. Did you mean $x^2 + 6x - 8$?
– N. F. Taussig
Nov 23 at 19:50
you can use this online graphing software to graph your function. desmos.com/calculator
– user25406
Nov 23 at 19:52
add a comment |
Do have any guidelines? Sketching without a computer involves using derivatives (first and second).
– Vasya
Nov 23 at 19:44
1
Denominator is $2(x+3)(x+2)$. This at least tells you the function is undefined at $x=-3, x=-2$. (We want to know the domain of the function).
– amWhy
Nov 23 at 19:48
3
$x^2 + 6x - 8x = x^2 - 2x$. Did you mean $x^2 + 6x - 8$?
– N. F. Taussig
Nov 23 at 19:50
you can use this online graphing software to graph your function. desmos.com/calculator
– user25406
Nov 23 at 19:52
Do have any guidelines? Sketching without a computer involves using derivatives (first and second).
– Vasya
Nov 23 at 19:44
Do have any guidelines? Sketching without a computer involves using derivatives (first and second).
– Vasya
Nov 23 at 19:44
1
1
Denominator is $2(x+3)(x+2)$. This at least tells you the function is undefined at $x=-3, x=-2$. (We want to know the domain of the function).
– amWhy
Nov 23 at 19:48
Denominator is $2(x+3)(x+2)$. This at least tells you the function is undefined at $x=-3, x=-2$. (We want to know the domain of the function).
– amWhy
Nov 23 at 19:48
3
3
$x^2 + 6x - 8x = x^2 - 2x$. Did you mean $x^2 + 6x - 8$?
– N. F. Taussig
Nov 23 at 19:50
$x^2 + 6x - 8x = x^2 - 2x$. Did you mean $x^2 + 6x - 8$?
– N. F. Taussig
Nov 23 at 19:50
you can use this online graphing software to graph your function. desmos.com/calculator
– user25406
Nov 23 at 19:52
you can use this online graphing software to graph your function. desmos.com/calculator
– user25406
Nov 23 at 19:52
add a comment |
5 Answers
5
active
oldest
votes
up vote
0
down vote
The x-intercepts are the points $x$ where $r(x)=0$. To find these points set the equation equal to 0 and solve for $x$:
$$frac{x^2+6x-8x}{2x^2+10x+12}=0$$
This is 0 when the numerator is 0, so solve the quadratic $x^2+6x-8x=0$ to obtain the x-intercepts.
The y-intercept is the point where the function $r$ crosses the y-axis. This is when $x=0$, so evaluate $r(0)$ to determine the y-intercept.
Because the polynomials in the numerator and denominator have the same degree, then the horizontal asymptote is the ratio of the leading coefficients for the numerator and denominator, in this case $1/2$ and the horizontal asymptote is at $y=1/2$.
The vertical asymptotes occur when the denominator obtains the value 0. So set the quadratic in the denominator equal to 0 and solve for $x$, these values of $x$ are the vertical asymptotes.
add a comment |
up vote
0
down vote
Why not simply plot it?
The zeros are when $x^2 + 6 x - 8 = 0$ and the divergences are when $2 x^2 + 10 x + 12 = 0$, and the limit is $x^2/(2 x^2) = 1/2$.
add a comment |
up vote
0
down vote
First, eliminate any common root of the numerator and denominator (only for finding the vertical asymptotes). Then find the roots of the denominator. They are all candidates for vertical asymptotes. Also find the limit of the function for $xto pm infty$ to find horizontal asymptotes. To find X-intercepts find $x$s for which $f(x)=0$ and similarly, Y-intercept can be found by substituting $x=0$ into the equation of the function. Clarifying all of this we have $$f(x)={x^2+6x-8over 2(x+2)(x+3)}$$hence, $x=-2$ and $x=-3$ are the vertical asymptotes since they are the roots of the denominator (and not that of the numerator). By tending $xto pminfty$ we also have $y={1over 2}$ as our only horizontal asymptote. The only Y-intercept is $$(0,-{2over 3})$$and the X-intercepts are $$x=-3pm sqrt{17}$$here is a sketch of the function
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
add a comment |
up vote
0
down vote
Getting there
$$
begin{array}{c|c|c|c}
x & x^2 + 6x-8 & 2 x^2 + 10 x + 12 & frac{x^2 + 6x-8}{2 x^2 + 10 x + 12} \ hline
-15&127&312&0.407 \
-14&104&264&0.3939 \
-13&83&220&0.3773 \
-12&64&180&0.3555 \
-11&47&144&0.3264 \
-10&32&112&0.2857 \
-9&19&84&0.2262 \
-8&8&60&0.1333 \
-7&-1&40&-0.025 \
-6&-8&24&-0.3333 \
-5&-13&12& -1.0833 \
-4&-16&4& -4 \
-3.75&-16.4375&2.625&-6.2619 \
-3.5&-16.75&1.5&-11.1667 \
-3.25&-16.9375&0.625&-27.1 \
-3&-17&0& mbox{vert asymp} \
-2.875&-16.984375&-0.21875&77.6428 \
-2.75&-16.9375&-0.375&45.1667 \
-2.625&-16.859375&-0.46875&35.9667 \
-2.5&-16.75&-0.5&33.5 \
-2.375&-16.609375&-0.46875&35.4333 \
-2.25&-16.4375&-0.375&43.8333 \
-2.125&-16.2343&-0.21875&74.2142 \
-2&-16&0& mbox{vert asymp} \
-1.75&-15.4375&0.625&-24.7 \
-1.5&-14.75&1.5&-9.8333 \
-1.25&-13.9375&2.625&-5.3095 \
-1&-13&4&-3.25 \
0&-8&12&-0.6667 \
1&-1&24&-0.0417 \
2&8&40&0.2 \
3&19&60&0.3167 \
4&32&84&0.3809 \
5&47&112&0.4196 \
6&64&144&0.4444 \
7&83&180&0.4611 \
end{array}
$$
add a comment |
up vote
0
down vote
To find the $x$-intercept, or the roots, set $y = 0$.
$$frac{x^2+6x-8x}{2x^2+10x+12} = 0$$
$$x^2+6x-8x = 0$$
$$x^2-2x = 0$$
$$x(x-2) = 0$$
$$x = 0 quad x = 2$$
To find the $y$-intercept, set $x = 0$.
$$frac{0^2+6(0)-8(0)}{2(0)^2+10(0)+12} = frac{0}{12} = 0$$
To find the vertical asymptote, see which value(s) of $x$ result in an undefined output. Set the denominator equal to $0$.
$$2x^2+10x+12 = 0$$
$$x^2+5x+6 = 0$$
$$(x+3)(x+2) = 0$$
$$x = -3 quad x = -2$$
To find the horizontal asymptote, find the limit to infinity. It becomes clear that as $x$ grows unboundedly large, the quadratic terms dominate. Both the numerator and denominator have leading quadratic terms, so the ratio of their coefficients will give the answer.
$$lim_{x to infty} frac{x^2+6x-8x}{2x^2+10x+12} = frac{x^2}{2x^2} = frac{1}{2}$$
Edit: If you meant $x^2+6x-8$ as the numerator (which you most probably did), you still carry out the same steps. For the $x$-intercepts, you reach
$$x^2+6x-8 = 0$$
which can’t be factored nicely, so just use the Quadratic Formula.
$$x = frac{-6pmsqrt{6^2-4(1)(-8)}}{2(1)} = frac{-6pmsqrt{68}}{2} = -3pmsqrt{17}$$
For the y-intercept, you get
$$frac{0^2+6(0)-8}{2(0)^2+10(0)+12} = frac{-8}{12} = -frac{2}{3}$$
The vertical and horizontal asympotes remain the same.
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The x-intercepts are the points $x$ where $r(x)=0$. To find these points set the equation equal to 0 and solve for $x$:
$$frac{x^2+6x-8x}{2x^2+10x+12}=0$$
This is 0 when the numerator is 0, so solve the quadratic $x^2+6x-8x=0$ to obtain the x-intercepts.
The y-intercept is the point where the function $r$ crosses the y-axis. This is when $x=0$, so evaluate $r(0)$ to determine the y-intercept.
Because the polynomials in the numerator and denominator have the same degree, then the horizontal asymptote is the ratio of the leading coefficients for the numerator and denominator, in this case $1/2$ and the horizontal asymptote is at $y=1/2$.
The vertical asymptotes occur when the denominator obtains the value 0. So set the quadratic in the denominator equal to 0 and solve for $x$, these values of $x$ are the vertical asymptotes.
add a comment |
up vote
0
down vote
The x-intercepts are the points $x$ where $r(x)=0$. To find these points set the equation equal to 0 and solve for $x$:
$$frac{x^2+6x-8x}{2x^2+10x+12}=0$$
This is 0 when the numerator is 0, so solve the quadratic $x^2+6x-8x=0$ to obtain the x-intercepts.
The y-intercept is the point where the function $r$ crosses the y-axis. This is when $x=0$, so evaluate $r(0)$ to determine the y-intercept.
Because the polynomials in the numerator and denominator have the same degree, then the horizontal asymptote is the ratio of the leading coefficients for the numerator and denominator, in this case $1/2$ and the horizontal asymptote is at $y=1/2$.
The vertical asymptotes occur when the denominator obtains the value 0. So set the quadratic in the denominator equal to 0 and solve for $x$, these values of $x$ are the vertical asymptotes.
add a comment |
up vote
0
down vote
up vote
0
down vote
The x-intercepts are the points $x$ where $r(x)=0$. To find these points set the equation equal to 0 and solve for $x$:
$$frac{x^2+6x-8x}{2x^2+10x+12}=0$$
This is 0 when the numerator is 0, so solve the quadratic $x^2+6x-8x=0$ to obtain the x-intercepts.
The y-intercept is the point where the function $r$ crosses the y-axis. This is when $x=0$, so evaluate $r(0)$ to determine the y-intercept.
Because the polynomials in the numerator and denominator have the same degree, then the horizontal asymptote is the ratio of the leading coefficients for the numerator and denominator, in this case $1/2$ and the horizontal asymptote is at $y=1/2$.
The vertical asymptotes occur when the denominator obtains the value 0. So set the quadratic in the denominator equal to 0 and solve for $x$, these values of $x$ are the vertical asymptotes.
The x-intercepts are the points $x$ where $r(x)=0$. To find these points set the equation equal to 0 and solve for $x$:
$$frac{x^2+6x-8x}{2x^2+10x+12}=0$$
This is 0 when the numerator is 0, so solve the quadratic $x^2+6x-8x=0$ to obtain the x-intercepts.
The y-intercept is the point where the function $r$ crosses the y-axis. This is when $x=0$, so evaluate $r(0)$ to determine the y-intercept.
Because the polynomials in the numerator and denominator have the same degree, then the horizontal asymptote is the ratio of the leading coefficients for the numerator and denominator, in this case $1/2$ and the horizontal asymptote is at $y=1/2$.
The vertical asymptotes occur when the denominator obtains the value 0. So set the quadratic in the denominator equal to 0 and solve for $x$, these values of $x$ are the vertical asymptotes.
answered Nov 23 at 19:43
gd1035
461210
461210
add a comment |
add a comment |
up vote
0
down vote
Why not simply plot it?
The zeros are when $x^2 + 6 x - 8 = 0$ and the divergences are when $2 x^2 + 10 x + 12 = 0$, and the limit is $x^2/(2 x^2) = 1/2$.
add a comment |
up vote
0
down vote
Why not simply plot it?
The zeros are when $x^2 + 6 x - 8 = 0$ and the divergences are when $2 x^2 + 10 x + 12 = 0$, and the limit is $x^2/(2 x^2) = 1/2$.
add a comment |
up vote
0
down vote
up vote
0
down vote
Why not simply plot it?
The zeros are when $x^2 + 6 x - 8 = 0$ and the divergences are when $2 x^2 + 10 x + 12 = 0$, and the limit is $x^2/(2 x^2) = 1/2$.
Why not simply plot it?
The zeros are when $x^2 + 6 x - 8 = 0$ and the divergences are when $2 x^2 + 10 x + 12 = 0$, and the limit is $x^2/(2 x^2) = 1/2$.
answered Nov 23 at 20:12
David G. Stork
9,54721232
9,54721232
add a comment |
add a comment |
up vote
0
down vote
First, eliminate any common root of the numerator and denominator (only for finding the vertical asymptotes). Then find the roots of the denominator. They are all candidates for vertical asymptotes. Also find the limit of the function for $xto pm infty$ to find horizontal asymptotes. To find X-intercepts find $x$s for which $f(x)=0$ and similarly, Y-intercept can be found by substituting $x=0$ into the equation of the function. Clarifying all of this we have $$f(x)={x^2+6x-8over 2(x+2)(x+3)}$$hence, $x=-2$ and $x=-3$ are the vertical asymptotes since they are the roots of the denominator (and not that of the numerator). By tending $xto pminfty$ we also have $y={1over 2}$ as our only horizontal asymptote. The only Y-intercept is $$(0,-{2over 3})$$and the X-intercepts are $$x=-3pm sqrt{17}$$here is a sketch of the function
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
add a comment |
up vote
0
down vote
First, eliminate any common root of the numerator and denominator (only for finding the vertical asymptotes). Then find the roots of the denominator. They are all candidates for vertical asymptotes. Also find the limit of the function for $xto pm infty$ to find horizontal asymptotes. To find X-intercepts find $x$s for which $f(x)=0$ and similarly, Y-intercept can be found by substituting $x=0$ into the equation of the function. Clarifying all of this we have $$f(x)={x^2+6x-8over 2(x+2)(x+3)}$$hence, $x=-2$ and $x=-3$ are the vertical asymptotes since they are the roots of the denominator (and not that of the numerator). By tending $xto pminfty$ we also have $y={1over 2}$ as our only horizontal asymptote. The only Y-intercept is $$(0,-{2over 3})$$and the X-intercepts are $$x=-3pm sqrt{17}$$here is a sketch of the function
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
add a comment |
up vote
0
down vote
up vote
0
down vote
First, eliminate any common root of the numerator and denominator (only for finding the vertical asymptotes). Then find the roots of the denominator. They are all candidates for vertical asymptotes. Also find the limit of the function for $xto pm infty$ to find horizontal asymptotes. To find X-intercepts find $x$s for which $f(x)=0$ and similarly, Y-intercept can be found by substituting $x=0$ into the equation of the function. Clarifying all of this we have $$f(x)={x^2+6x-8over 2(x+2)(x+3)}$$hence, $x=-2$ and $x=-3$ are the vertical asymptotes since they are the roots of the denominator (and not that of the numerator). By tending $xto pminfty$ we also have $y={1over 2}$ as our only horizontal asymptote. The only Y-intercept is $$(0,-{2over 3})$$and the X-intercepts are $$x=-3pm sqrt{17}$$here is a sketch of the function
First, eliminate any common root of the numerator and denominator (only for finding the vertical asymptotes). Then find the roots of the denominator. They are all candidates for vertical asymptotes. Also find the limit of the function for $xto pm infty$ to find horizontal asymptotes. To find X-intercepts find $x$s for which $f(x)=0$ and similarly, Y-intercept can be found by substituting $x=0$ into the equation of the function. Clarifying all of this we have $$f(x)={x^2+6x-8over 2(x+2)(x+3)}$$hence, $x=-2$ and $x=-3$ are the vertical asymptotes since they are the roots of the denominator (and not that of the numerator). By tending $xto pminfty$ we also have $y={1over 2}$ as our only horizontal asymptote. The only Y-intercept is $$(0,-{2over 3})$$and the X-intercepts are $$x=-3pm sqrt{17}$$here is a sketch of the function
answered Nov 23 at 20:41
Mostafa Ayaz
13.6k3836
13.6k3836
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
add a comment |
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
Thank you so much for the detailed answer! I really appreciate it! I was wondering if I could send you an email? If not, that's cool.
– User231
Nov 27 at 5:17
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
You're welcome. Sure you can. If you want to stay in touch, here is my email address ayazmostafa@gmail.com
– Mostafa Ayaz
Nov 27 at 8:00
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
I sent you an email! I mentioned that it's me in the subject.
– User231
Nov 28 at 1:01
add a comment |
up vote
0
down vote
Getting there
$$
begin{array}{c|c|c|c}
x & x^2 + 6x-8 & 2 x^2 + 10 x + 12 & frac{x^2 + 6x-8}{2 x^2 + 10 x + 12} \ hline
-15&127&312&0.407 \
-14&104&264&0.3939 \
-13&83&220&0.3773 \
-12&64&180&0.3555 \
-11&47&144&0.3264 \
-10&32&112&0.2857 \
-9&19&84&0.2262 \
-8&8&60&0.1333 \
-7&-1&40&-0.025 \
-6&-8&24&-0.3333 \
-5&-13&12& -1.0833 \
-4&-16&4& -4 \
-3.75&-16.4375&2.625&-6.2619 \
-3.5&-16.75&1.5&-11.1667 \
-3.25&-16.9375&0.625&-27.1 \
-3&-17&0& mbox{vert asymp} \
-2.875&-16.984375&-0.21875&77.6428 \
-2.75&-16.9375&-0.375&45.1667 \
-2.625&-16.859375&-0.46875&35.9667 \
-2.5&-16.75&-0.5&33.5 \
-2.375&-16.609375&-0.46875&35.4333 \
-2.25&-16.4375&-0.375&43.8333 \
-2.125&-16.2343&-0.21875&74.2142 \
-2&-16&0& mbox{vert asymp} \
-1.75&-15.4375&0.625&-24.7 \
-1.5&-14.75&1.5&-9.8333 \
-1.25&-13.9375&2.625&-5.3095 \
-1&-13&4&-3.25 \
0&-8&12&-0.6667 \
1&-1&24&-0.0417 \
2&8&40&0.2 \
3&19&60&0.3167 \
4&32&84&0.3809 \
5&47&112&0.4196 \
6&64&144&0.4444 \
7&83&180&0.4611 \
end{array}
$$
add a comment |
up vote
0
down vote
Getting there
$$
begin{array}{c|c|c|c}
x & x^2 + 6x-8 & 2 x^2 + 10 x + 12 & frac{x^2 + 6x-8}{2 x^2 + 10 x + 12} \ hline
-15&127&312&0.407 \
-14&104&264&0.3939 \
-13&83&220&0.3773 \
-12&64&180&0.3555 \
-11&47&144&0.3264 \
-10&32&112&0.2857 \
-9&19&84&0.2262 \
-8&8&60&0.1333 \
-7&-1&40&-0.025 \
-6&-8&24&-0.3333 \
-5&-13&12& -1.0833 \
-4&-16&4& -4 \
-3.75&-16.4375&2.625&-6.2619 \
-3.5&-16.75&1.5&-11.1667 \
-3.25&-16.9375&0.625&-27.1 \
-3&-17&0& mbox{vert asymp} \
-2.875&-16.984375&-0.21875&77.6428 \
-2.75&-16.9375&-0.375&45.1667 \
-2.625&-16.859375&-0.46875&35.9667 \
-2.5&-16.75&-0.5&33.5 \
-2.375&-16.609375&-0.46875&35.4333 \
-2.25&-16.4375&-0.375&43.8333 \
-2.125&-16.2343&-0.21875&74.2142 \
-2&-16&0& mbox{vert asymp} \
-1.75&-15.4375&0.625&-24.7 \
-1.5&-14.75&1.5&-9.8333 \
-1.25&-13.9375&2.625&-5.3095 \
-1&-13&4&-3.25 \
0&-8&12&-0.6667 \
1&-1&24&-0.0417 \
2&8&40&0.2 \
3&19&60&0.3167 \
4&32&84&0.3809 \
5&47&112&0.4196 \
6&64&144&0.4444 \
7&83&180&0.4611 \
end{array}
$$
add a comment |
up vote
0
down vote
up vote
0
down vote
Getting there
$$
begin{array}{c|c|c|c}
x & x^2 + 6x-8 & 2 x^2 + 10 x + 12 & frac{x^2 + 6x-8}{2 x^2 + 10 x + 12} \ hline
-15&127&312&0.407 \
-14&104&264&0.3939 \
-13&83&220&0.3773 \
-12&64&180&0.3555 \
-11&47&144&0.3264 \
-10&32&112&0.2857 \
-9&19&84&0.2262 \
-8&8&60&0.1333 \
-7&-1&40&-0.025 \
-6&-8&24&-0.3333 \
-5&-13&12& -1.0833 \
-4&-16&4& -4 \
-3.75&-16.4375&2.625&-6.2619 \
-3.5&-16.75&1.5&-11.1667 \
-3.25&-16.9375&0.625&-27.1 \
-3&-17&0& mbox{vert asymp} \
-2.875&-16.984375&-0.21875&77.6428 \
-2.75&-16.9375&-0.375&45.1667 \
-2.625&-16.859375&-0.46875&35.9667 \
-2.5&-16.75&-0.5&33.5 \
-2.375&-16.609375&-0.46875&35.4333 \
-2.25&-16.4375&-0.375&43.8333 \
-2.125&-16.2343&-0.21875&74.2142 \
-2&-16&0& mbox{vert asymp} \
-1.75&-15.4375&0.625&-24.7 \
-1.5&-14.75&1.5&-9.8333 \
-1.25&-13.9375&2.625&-5.3095 \
-1&-13&4&-3.25 \
0&-8&12&-0.6667 \
1&-1&24&-0.0417 \
2&8&40&0.2 \
3&19&60&0.3167 \
4&32&84&0.3809 \
5&47&112&0.4196 \
6&64&144&0.4444 \
7&83&180&0.4611 \
end{array}
$$
Getting there
$$
begin{array}{c|c|c|c}
x & x^2 + 6x-8 & 2 x^2 + 10 x + 12 & frac{x^2 + 6x-8}{2 x^2 + 10 x + 12} \ hline
-15&127&312&0.407 \
-14&104&264&0.3939 \
-13&83&220&0.3773 \
-12&64&180&0.3555 \
-11&47&144&0.3264 \
-10&32&112&0.2857 \
-9&19&84&0.2262 \
-8&8&60&0.1333 \
-7&-1&40&-0.025 \
-6&-8&24&-0.3333 \
-5&-13&12& -1.0833 \
-4&-16&4& -4 \
-3.75&-16.4375&2.625&-6.2619 \
-3.5&-16.75&1.5&-11.1667 \
-3.25&-16.9375&0.625&-27.1 \
-3&-17&0& mbox{vert asymp} \
-2.875&-16.984375&-0.21875&77.6428 \
-2.75&-16.9375&-0.375&45.1667 \
-2.625&-16.859375&-0.46875&35.9667 \
-2.5&-16.75&-0.5&33.5 \
-2.375&-16.609375&-0.46875&35.4333 \
-2.25&-16.4375&-0.375&43.8333 \
-2.125&-16.2343&-0.21875&74.2142 \
-2&-16&0& mbox{vert asymp} \
-1.75&-15.4375&0.625&-24.7 \
-1.5&-14.75&1.5&-9.8333 \
-1.25&-13.9375&2.625&-5.3095 \
-1&-13&4&-3.25 \
0&-8&12&-0.6667 \
1&-1&24&-0.0417 \
2&8&40&0.2 \
3&19&60&0.3167 \
4&32&84&0.3809 \
5&47&112&0.4196 \
6&64&144&0.4444 \
7&83&180&0.4611 \
end{array}
$$
edited Nov 23 at 20:45
answered Nov 23 at 20:22
Will Jagy
101k598198
101k598198
add a comment |
add a comment |
up vote
0
down vote
To find the $x$-intercept, or the roots, set $y = 0$.
$$frac{x^2+6x-8x}{2x^2+10x+12} = 0$$
$$x^2+6x-8x = 0$$
$$x^2-2x = 0$$
$$x(x-2) = 0$$
$$x = 0 quad x = 2$$
To find the $y$-intercept, set $x = 0$.
$$frac{0^2+6(0)-8(0)}{2(0)^2+10(0)+12} = frac{0}{12} = 0$$
To find the vertical asymptote, see which value(s) of $x$ result in an undefined output. Set the denominator equal to $0$.
$$2x^2+10x+12 = 0$$
$$x^2+5x+6 = 0$$
$$(x+3)(x+2) = 0$$
$$x = -3 quad x = -2$$
To find the horizontal asymptote, find the limit to infinity. It becomes clear that as $x$ grows unboundedly large, the quadratic terms dominate. Both the numerator and denominator have leading quadratic terms, so the ratio of their coefficients will give the answer.
$$lim_{x to infty} frac{x^2+6x-8x}{2x^2+10x+12} = frac{x^2}{2x^2} = frac{1}{2}$$
Edit: If you meant $x^2+6x-8$ as the numerator (which you most probably did), you still carry out the same steps. For the $x$-intercepts, you reach
$$x^2+6x-8 = 0$$
which can’t be factored nicely, so just use the Quadratic Formula.
$$x = frac{-6pmsqrt{6^2-4(1)(-8)}}{2(1)} = frac{-6pmsqrt{68}}{2} = -3pmsqrt{17}$$
For the y-intercept, you get
$$frac{0^2+6(0)-8}{2(0)^2+10(0)+12} = frac{-8}{12} = -frac{2}{3}$$
The vertical and horizontal asympotes remain the same.
add a comment |
up vote
0
down vote
To find the $x$-intercept, or the roots, set $y = 0$.
$$frac{x^2+6x-8x}{2x^2+10x+12} = 0$$
$$x^2+6x-8x = 0$$
$$x^2-2x = 0$$
$$x(x-2) = 0$$
$$x = 0 quad x = 2$$
To find the $y$-intercept, set $x = 0$.
$$frac{0^2+6(0)-8(0)}{2(0)^2+10(0)+12} = frac{0}{12} = 0$$
To find the vertical asymptote, see which value(s) of $x$ result in an undefined output. Set the denominator equal to $0$.
$$2x^2+10x+12 = 0$$
$$x^2+5x+6 = 0$$
$$(x+3)(x+2) = 0$$
$$x = -3 quad x = -2$$
To find the horizontal asymptote, find the limit to infinity. It becomes clear that as $x$ grows unboundedly large, the quadratic terms dominate. Both the numerator and denominator have leading quadratic terms, so the ratio of their coefficients will give the answer.
$$lim_{x to infty} frac{x^2+6x-8x}{2x^2+10x+12} = frac{x^2}{2x^2} = frac{1}{2}$$
Edit: If you meant $x^2+6x-8$ as the numerator (which you most probably did), you still carry out the same steps. For the $x$-intercepts, you reach
$$x^2+6x-8 = 0$$
which can’t be factored nicely, so just use the Quadratic Formula.
$$x = frac{-6pmsqrt{6^2-4(1)(-8)}}{2(1)} = frac{-6pmsqrt{68}}{2} = -3pmsqrt{17}$$
For the y-intercept, you get
$$frac{0^2+6(0)-8}{2(0)^2+10(0)+12} = frac{-8}{12} = -frac{2}{3}$$
The vertical and horizontal asympotes remain the same.
add a comment |
up vote
0
down vote
up vote
0
down vote
To find the $x$-intercept, or the roots, set $y = 0$.
$$frac{x^2+6x-8x}{2x^2+10x+12} = 0$$
$$x^2+6x-8x = 0$$
$$x^2-2x = 0$$
$$x(x-2) = 0$$
$$x = 0 quad x = 2$$
To find the $y$-intercept, set $x = 0$.
$$frac{0^2+6(0)-8(0)}{2(0)^2+10(0)+12} = frac{0}{12} = 0$$
To find the vertical asymptote, see which value(s) of $x$ result in an undefined output. Set the denominator equal to $0$.
$$2x^2+10x+12 = 0$$
$$x^2+5x+6 = 0$$
$$(x+3)(x+2) = 0$$
$$x = -3 quad x = -2$$
To find the horizontal asymptote, find the limit to infinity. It becomes clear that as $x$ grows unboundedly large, the quadratic terms dominate. Both the numerator and denominator have leading quadratic terms, so the ratio of their coefficients will give the answer.
$$lim_{x to infty} frac{x^2+6x-8x}{2x^2+10x+12} = frac{x^2}{2x^2} = frac{1}{2}$$
Edit: If you meant $x^2+6x-8$ as the numerator (which you most probably did), you still carry out the same steps. For the $x$-intercepts, you reach
$$x^2+6x-8 = 0$$
which can’t be factored nicely, so just use the Quadratic Formula.
$$x = frac{-6pmsqrt{6^2-4(1)(-8)}}{2(1)} = frac{-6pmsqrt{68}}{2} = -3pmsqrt{17}$$
For the y-intercept, you get
$$frac{0^2+6(0)-8}{2(0)^2+10(0)+12} = frac{-8}{12} = -frac{2}{3}$$
The vertical and horizontal asympotes remain the same.
To find the $x$-intercept, or the roots, set $y = 0$.
$$frac{x^2+6x-8x}{2x^2+10x+12} = 0$$
$$x^2+6x-8x = 0$$
$$x^2-2x = 0$$
$$x(x-2) = 0$$
$$x = 0 quad x = 2$$
To find the $y$-intercept, set $x = 0$.
$$frac{0^2+6(0)-8(0)}{2(0)^2+10(0)+12} = frac{0}{12} = 0$$
To find the vertical asymptote, see which value(s) of $x$ result in an undefined output. Set the denominator equal to $0$.
$$2x^2+10x+12 = 0$$
$$x^2+5x+6 = 0$$
$$(x+3)(x+2) = 0$$
$$x = -3 quad x = -2$$
To find the horizontal asymptote, find the limit to infinity. It becomes clear that as $x$ grows unboundedly large, the quadratic terms dominate. Both the numerator and denominator have leading quadratic terms, so the ratio of their coefficients will give the answer.
$$lim_{x to infty} frac{x^2+6x-8x}{2x^2+10x+12} = frac{x^2}{2x^2} = frac{1}{2}$$
Edit: If you meant $x^2+6x-8$ as the numerator (which you most probably did), you still carry out the same steps. For the $x$-intercepts, you reach
$$x^2+6x-8 = 0$$
which can’t be factored nicely, so just use the Quadratic Formula.
$$x = frac{-6pmsqrt{6^2-4(1)(-8)}}{2(1)} = frac{-6pmsqrt{68}}{2} = -3pmsqrt{17}$$
For the y-intercept, you get
$$frac{0^2+6(0)-8}{2(0)^2+10(0)+12} = frac{-8}{12} = -frac{2}{3}$$
The vertical and horizontal asympotes remain the same.
edited Nov 23 at 21:48
answered Nov 23 at 19:51
KM101
3,945417
3,945417
add a comment |
add a comment |
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Do have any guidelines? Sketching without a computer involves using derivatives (first and second).
– Vasya
Nov 23 at 19:44
1
Denominator is $2(x+3)(x+2)$. This at least tells you the function is undefined at $x=-3, x=-2$. (We want to know the domain of the function).
– amWhy
Nov 23 at 19:48
3
$x^2 + 6x - 8x = x^2 - 2x$. Did you mean $x^2 + 6x - 8$?
– N. F. Taussig
Nov 23 at 19:50
you can use this online graphing software to graph your function. desmos.com/calculator
– user25406
Nov 23 at 19:52