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How do I graph this?
How to solve this 4 terms equation?How to solve exponential equations?How do I solve the chessboard problem?How to solve this linear equation $x:y:3=2:1:5$How do you graph a negative number raised to $x$?Finding a point on a relationship without trial and error.Finding The Complex Roots Of A PolynomialWhy Can A Graph Pass Through Its Horizontal Asymptote?Can someone please explain to me why cotangent graphs look the way they do?How do people come up with solutions like this?
$begingroup$
I am just a high school student, so please excuse me if this question sound silly to you.
How do I graph this : $dfrac9x - dfrac4y = 8$
I could do trial and error, but is there a more systematic way that I can use to graph this equation.
algebra-precalculus graphing-functions
$endgroup$
add a comment |
$begingroup$
I am just a high school student, so please excuse me if this question sound silly to you.
How do I graph this : $dfrac9x - dfrac4y = 8$
I could do trial and error, but is there a more systematic way that I can use to graph this equation.
algebra-precalculus graphing-functions
$endgroup$
$begingroup$
Can you solve for $y$ ?
$endgroup$
– dmtri
yesterday
$begingroup$
Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions?
$endgroup$
– Rory Daulton
yesterday
add a comment |
$begingroup$
I am just a high school student, so please excuse me if this question sound silly to you.
How do I graph this : $dfrac9x - dfrac4y = 8$
I could do trial and error, but is there a more systematic way that I can use to graph this equation.
algebra-precalculus graphing-functions
$endgroup$
I am just a high school student, so please excuse me if this question sound silly to you.
How do I graph this : $dfrac9x - dfrac4y = 8$
I could do trial and error, but is there a more systematic way that I can use to graph this equation.
algebra-precalculus graphing-functions
algebra-precalculus graphing-functions
edited yesterday
ArsenBerk
7,58531338
7,58531338
asked yesterday
Aditya BharadwajAditya Bharadwaj
1211
1211
$begingroup$
Can you solve for $y$ ?
$endgroup$
– dmtri
yesterday
$begingroup$
Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions?
$endgroup$
– Rory Daulton
yesterday
add a comment |
$begingroup$
Can you solve for $y$ ?
$endgroup$
– dmtri
yesterday
$begingroup$
Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions?
$endgroup$
– Rory Daulton
yesterday
$begingroup$
Can you solve for $y$ ?
$endgroup$
– dmtri
yesterday
$begingroup$
Can you solve for $y$ ?
$endgroup$
– dmtri
yesterday
$begingroup$
Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions?
$endgroup$
– Rory Daulton
yesterday
$begingroup$
Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions?
$endgroup$
– Rory Daulton
yesterday
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I would write $$frac9x-8=frac4y$$ so,
$$frac9-8xx=frac4y$$ and so, $$y=frac4x9-8x$$ and then use a function plotter from the internet.
$endgroup$
add a comment |
$begingroup$
I'll assume you want to graph that equation without a graphing calculator or graphing program, and that you want to use only pre-calculus techniques.
So we want to transform the equation into something resembling one of the "standard equations" that we have seen before. First, we simplify the equation by removing the fractions and moving all terms to one side of the equation.
$$frac 9x - frac 4y = 8$$
$$left(frac 9x - frac 4yright)xy = 8xy$$
$$9y - 4x = 8xy$$
$$8xy + 4x - 9y = 0$$
That left side looks factorable. We'll make the first coefficient into one.
$$frac 18(8xy + 4x - 9y)= frac 18cdot 0$$
$$xy + frac 12x - frac 98y = 0$$
Now that looks much like factoring a polynomial. We see that we can split the $frac 12$ and the $-frac 98$ by making the left side into the product of two binomials. But first we need to add the appropriate constant term.
$$xy + frac 12x - frac 98y + -frac 98cdotfrac 12 = -frac 98cdotfrac 12$$
$$left(x-frac 98right) left(y+frac 12right) = -left(frac 34right)^2$$
Now we see this is a transformation of the graph of the standard equation
$$xy=1$$
This is a hyperbola. Our graph has a negative number on the right, so it is reflected in an axis, so the branches of the hyperbola are in the upper-left and lower-right areas of the graph. The right side is not one so the vertices of the hyperbola have an x-distance and a y-distance of $frac 34$ from the center. The graph is translated so the vertical asymptote is the line $x=frac 98$ and the horizontal asymptote is the line $y=-frac 12$.
One last thing. We started all this by multiplying by $xy$ to remove the fractions. That means that we must pay attention to the points where $x=0$ or $y=0$. The original equation shows that such points are not on the graph. So even though the origin is included in our final equation, it is not on the original graph. So our hyperbola has a hole at the origin.
That information makes it easy to sketch the graph. We can confirm this by graphing one of the equations on a graphing calculator. Here is $8xy+4x-9y=0$ on the TI-Nspire CX scratchpad screen. Except for the hole at the origin, this agrees with my analysis.
$endgroup$
add a comment |
$begingroup$
Yes, there is. Note thatbeginalignfrac9x-frac4y=8&ifffrac4y=frac9x-8\&iff y=frac4frac9x-8=frac4x9-8x=-frac12+frac918-16x.endalignSo, graph the function $xmapsto-dfrac12+dfrac918-16x$.
$endgroup$
add a comment |
$begingroup$
Notice the coefficients of $frac1x$ and $frac1y$, we can alter $x$ in multiples of $3$ or $9$ to get corresponding values of $y$. Also $x neq0$, and $yneq 0$, to satisfy the equation:
$$frac9x-frac4y=8$$
Rewrite the equation as :
$$y=frac4frac9x-8$$ and just put in values of $x$ as $pm1,pm2,pm3,pm9,pm12,pm15,pm18$ as well as $pm frac12,pm frac13, pm frac19, pm frac112$ to get a rough idea of the curve.
Note there will be a break in the curve at $x=frac98$.
You can also compute the derivatives and verify in which part is the function increasing or decreasing. To plot it you can use MATLAB or matplotlib in python. Hope this helps....
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would write $$frac9x-8=frac4y$$ so,
$$frac9-8xx=frac4y$$ and so, $$y=frac4x9-8x$$ and then use a function plotter from the internet.
$endgroup$
add a comment |
$begingroup$
I would write $$frac9x-8=frac4y$$ so,
$$frac9-8xx=frac4y$$ and so, $$y=frac4x9-8x$$ and then use a function plotter from the internet.
$endgroup$
add a comment |
$begingroup$
I would write $$frac9x-8=frac4y$$ so,
$$frac9-8xx=frac4y$$ and so, $$y=frac4x9-8x$$ and then use a function plotter from the internet.
$endgroup$
I would write $$frac9x-8=frac4y$$ so,
$$frac9-8xx=frac4y$$ and so, $$y=frac4x9-8x$$ and then use a function plotter from the internet.
edited yesterday
SNEHIL SANYAL
618110
618110
answered yesterday
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
77.4k42866
add a comment |
add a comment |
$begingroup$
I'll assume you want to graph that equation without a graphing calculator or graphing program, and that you want to use only pre-calculus techniques.
So we want to transform the equation into something resembling one of the "standard equations" that we have seen before. First, we simplify the equation by removing the fractions and moving all terms to one side of the equation.
$$frac 9x - frac 4y = 8$$
$$left(frac 9x - frac 4yright)xy = 8xy$$
$$9y - 4x = 8xy$$
$$8xy + 4x - 9y = 0$$
That left side looks factorable. We'll make the first coefficient into one.
$$frac 18(8xy + 4x - 9y)= frac 18cdot 0$$
$$xy + frac 12x - frac 98y = 0$$
Now that looks much like factoring a polynomial. We see that we can split the $frac 12$ and the $-frac 98$ by making the left side into the product of two binomials. But first we need to add the appropriate constant term.
$$xy + frac 12x - frac 98y + -frac 98cdotfrac 12 = -frac 98cdotfrac 12$$
$$left(x-frac 98right) left(y+frac 12right) = -left(frac 34right)^2$$
Now we see this is a transformation of the graph of the standard equation
$$xy=1$$
This is a hyperbola. Our graph has a negative number on the right, so it is reflected in an axis, so the branches of the hyperbola are in the upper-left and lower-right areas of the graph. The right side is not one so the vertices of the hyperbola have an x-distance and a y-distance of $frac 34$ from the center. The graph is translated so the vertical asymptote is the line $x=frac 98$ and the horizontal asymptote is the line $y=-frac 12$.
One last thing. We started all this by multiplying by $xy$ to remove the fractions. That means that we must pay attention to the points where $x=0$ or $y=0$. The original equation shows that such points are not on the graph. So even though the origin is included in our final equation, it is not on the original graph. So our hyperbola has a hole at the origin.
That information makes it easy to sketch the graph. We can confirm this by graphing one of the equations on a graphing calculator. Here is $8xy+4x-9y=0$ on the TI-Nspire CX scratchpad screen. Except for the hole at the origin, this agrees with my analysis.
$endgroup$
add a comment |
$begingroup$
I'll assume you want to graph that equation without a graphing calculator or graphing program, and that you want to use only pre-calculus techniques.
So we want to transform the equation into something resembling one of the "standard equations" that we have seen before. First, we simplify the equation by removing the fractions and moving all terms to one side of the equation.
$$frac 9x - frac 4y = 8$$
$$left(frac 9x - frac 4yright)xy = 8xy$$
$$9y - 4x = 8xy$$
$$8xy + 4x - 9y = 0$$
That left side looks factorable. We'll make the first coefficient into one.
$$frac 18(8xy + 4x - 9y)= frac 18cdot 0$$
$$xy + frac 12x - frac 98y = 0$$
Now that looks much like factoring a polynomial. We see that we can split the $frac 12$ and the $-frac 98$ by making the left side into the product of two binomials. But first we need to add the appropriate constant term.
$$xy + frac 12x - frac 98y + -frac 98cdotfrac 12 = -frac 98cdotfrac 12$$
$$left(x-frac 98right) left(y+frac 12right) = -left(frac 34right)^2$$
Now we see this is a transformation of the graph of the standard equation
$$xy=1$$
This is a hyperbola. Our graph has a negative number on the right, so it is reflected in an axis, so the branches of the hyperbola are in the upper-left and lower-right areas of the graph. The right side is not one so the vertices of the hyperbola have an x-distance and a y-distance of $frac 34$ from the center. The graph is translated so the vertical asymptote is the line $x=frac 98$ and the horizontal asymptote is the line $y=-frac 12$.
One last thing. We started all this by multiplying by $xy$ to remove the fractions. That means that we must pay attention to the points where $x=0$ or $y=0$. The original equation shows that such points are not on the graph. So even though the origin is included in our final equation, it is not on the original graph. So our hyperbola has a hole at the origin.
That information makes it easy to sketch the graph. We can confirm this by graphing one of the equations on a graphing calculator. Here is $8xy+4x-9y=0$ on the TI-Nspire CX scratchpad screen. Except for the hole at the origin, this agrees with my analysis.
$endgroup$
add a comment |
$begingroup$
I'll assume you want to graph that equation without a graphing calculator or graphing program, and that you want to use only pre-calculus techniques.
So we want to transform the equation into something resembling one of the "standard equations" that we have seen before. First, we simplify the equation by removing the fractions and moving all terms to one side of the equation.
$$frac 9x - frac 4y = 8$$
$$left(frac 9x - frac 4yright)xy = 8xy$$
$$9y - 4x = 8xy$$
$$8xy + 4x - 9y = 0$$
That left side looks factorable. We'll make the first coefficient into one.
$$frac 18(8xy + 4x - 9y)= frac 18cdot 0$$
$$xy + frac 12x - frac 98y = 0$$
Now that looks much like factoring a polynomial. We see that we can split the $frac 12$ and the $-frac 98$ by making the left side into the product of two binomials. But first we need to add the appropriate constant term.
$$xy + frac 12x - frac 98y + -frac 98cdotfrac 12 = -frac 98cdotfrac 12$$
$$left(x-frac 98right) left(y+frac 12right) = -left(frac 34right)^2$$
Now we see this is a transformation of the graph of the standard equation
$$xy=1$$
This is a hyperbola. Our graph has a negative number on the right, so it is reflected in an axis, so the branches of the hyperbola are in the upper-left and lower-right areas of the graph. The right side is not one so the vertices of the hyperbola have an x-distance and a y-distance of $frac 34$ from the center. The graph is translated so the vertical asymptote is the line $x=frac 98$ and the horizontal asymptote is the line $y=-frac 12$.
One last thing. We started all this by multiplying by $xy$ to remove the fractions. That means that we must pay attention to the points where $x=0$ or $y=0$. The original equation shows that such points are not on the graph. So even though the origin is included in our final equation, it is not on the original graph. So our hyperbola has a hole at the origin.
That information makes it easy to sketch the graph. We can confirm this by graphing one of the equations on a graphing calculator. Here is $8xy+4x-9y=0$ on the TI-Nspire CX scratchpad screen. Except for the hole at the origin, this agrees with my analysis.
$endgroup$
I'll assume you want to graph that equation without a graphing calculator or graphing program, and that you want to use only pre-calculus techniques.
So we want to transform the equation into something resembling one of the "standard equations" that we have seen before. First, we simplify the equation by removing the fractions and moving all terms to one side of the equation.
$$frac 9x - frac 4y = 8$$
$$left(frac 9x - frac 4yright)xy = 8xy$$
$$9y - 4x = 8xy$$
$$8xy + 4x - 9y = 0$$
That left side looks factorable. We'll make the first coefficient into one.
$$frac 18(8xy + 4x - 9y)= frac 18cdot 0$$
$$xy + frac 12x - frac 98y = 0$$
Now that looks much like factoring a polynomial. We see that we can split the $frac 12$ and the $-frac 98$ by making the left side into the product of two binomials. But first we need to add the appropriate constant term.
$$xy + frac 12x - frac 98y + -frac 98cdotfrac 12 = -frac 98cdotfrac 12$$
$$left(x-frac 98right) left(y+frac 12right) = -left(frac 34right)^2$$
Now we see this is a transformation of the graph of the standard equation
$$xy=1$$
This is a hyperbola. Our graph has a negative number on the right, so it is reflected in an axis, so the branches of the hyperbola are in the upper-left and lower-right areas of the graph. The right side is not one so the vertices of the hyperbola have an x-distance and a y-distance of $frac 34$ from the center. The graph is translated so the vertical asymptote is the line $x=frac 98$ and the horizontal asymptote is the line $y=-frac 12$.
One last thing. We started all this by multiplying by $xy$ to remove the fractions. That means that we must pay attention to the points where $x=0$ or $y=0$. The original equation shows that such points are not on the graph. So even though the origin is included in our final equation, it is not on the original graph. So our hyperbola has a hole at the origin.
That information makes it easy to sketch the graph. We can confirm this by graphing one of the equations on a graphing calculator. Here is $8xy+4x-9y=0$ on the TI-Nspire CX scratchpad screen. Except for the hole at the origin, this agrees with my analysis.
edited yesterday
answered yesterday
Rory DaultonRory Daulton
29.5k63355
29.5k63355
add a comment |
add a comment |
$begingroup$
Yes, there is. Note thatbeginalignfrac9x-frac4y=8&ifffrac4y=frac9x-8\&iff y=frac4frac9x-8=frac4x9-8x=-frac12+frac918-16x.endalignSo, graph the function $xmapsto-dfrac12+dfrac918-16x$.
$endgroup$
add a comment |
$begingroup$
Yes, there is. Note thatbeginalignfrac9x-frac4y=8&ifffrac4y=frac9x-8\&iff y=frac4frac9x-8=frac4x9-8x=-frac12+frac918-16x.endalignSo, graph the function $xmapsto-dfrac12+dfrac918-16x$.
$endgroup$
add a comment |
$begingroup$
Yes, there is. Note thatbeginalignfrac9x-frac4y=8&ifffrac4y=frac9x-8\&iff y=frac4frac9x-8=frac4x9-8x=-frac12+frac918-16x.endalignSo, graph the function $xmapsto-dfrac12+dfrac918-16x$.
$endgroup$
Yes, there is. Note thatbeginalignfrac9x-frac4y=8&ifffrac4y=frac9x-8\&iff y=frac4frac9x-8=frac4x9-8x=-frac12+frac918-16x.endalignSo, graph the function $xmapsto-dfrac12+dfrac918-16x$.
answered yesterday
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
add a comment |
$begingroup$
Notice the coefficients of $frac1x$ and $frac1y$, we can alter $x$ in multiples of $3$ or $9$ to get corresponding values of $y$. Also $x neq0$, and $yneq 0$, to satisfy the equation:
$$frac9x-frac4y=8$$
Rewrite the equation as :
$$y=frac4frac9x-8$$ and just put in values of $x$ as $pm1,pm2,pm3,pm9,pm12,pm15,pm18$ as well as $pm frac12,pm frac13, pm frac19, pm frac112$ to get a rough idea of the curve.
Note there will be a break in the curve at $x=frac98$.
You can also compute the derivatives and verify in which part is the function increasing or decreasing. To plot it you can use MATLAB or matplotlib in python. Hope this helps....
$endgroup$
add a comment |
$begingroup$
Notice the coefficients of $frac1x$ and $frac1y$, we can alter $x$ in multiples of $3$ or $9$ to get corresponding values of $y$. Also $x neq0$, and $yneq 0$, to satisfy the equation:
$$frac9x-frac4y=8$$
Rewrite the equation as :
$$y=frac4frac9x-8$$ and just put in values of $x$ as $pm1,pm2,pm3,pm9,pm12,pm15,pm18$ as well as $pm frac12,pm frac13, pm frac19, pm frac112$ to get a rough idea of the curve.
Note there will be a break in the curve at $x=frac98$.
You can also compute the derivatives and verify in which part is the function increasing or decreasing. To plot it you can use MATLAB or matplotlib in python. Hope this helps....
$endgroup$
add a comment |
$begingroup$
Notice the coefficients of $frac1x$ and $frac1y$, we can alter $x$ in multiples of $3$ or $9$ to get corresponding values of $y$. Also $x neq0$, and $yneq 0$, to satisfy the equation:
$$frac9x-frac4y=8$$
Rewrite the equation as :
$$y=frac4frac9x-8$$ and just put in values of $x$ as $pm1,pm2,pm3,pm9,pm12,pm15,pm18$ as well as $pm frac12,pm frac13, pm frac19, pm frac112$ to get a rough idea of the curve.
Note there will be a break in the curve at $x=frac98$.
You can also compute the derivatives and verify in which part is the function increasing or decreasing. To plot it you can use MATLAB or matplotlib in python. Hope this helps....
$endgroup$
Notice the coefficients of $frac1x$ and $frac1y$, we can alter $x$ in multiples of $3$ or $9$ to get corresponding values of $y$. Also $x neq0$, and $yneq 0$, to satisfy the equation:
$$frac9x-frac4y=8$$
Rewrite the equation as :
$$y=frac4frac9x-8$$ and just put in values of $x$ as $pm1,pm2,pm3,pm9,pm12,pm15,pm18$ as well as $pm frac12,pm frac13, pm frac19, pm frac112$ to get a rough idea of the curve.
Note there will be a break in the curve at $x=frac98$.
You can also compute the derivatives and verify in which part is the function increasing or decreasing. To plot it you can use MATLAB or matplotlib in python. Hope this helps....
answered yesterday
SNEHIL SANYALSNEHIL SANYAL
618110
618110
add a comment |
add a comment |
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$begingroup$
Can you solve for $y$ ?
$endgroup$
– dmtri
yesterday
$begingroup$
Please explain what you mean by "graph this equation." Do you mean get it into a form so you can use a graphing calculator or program to show the graph, or do you mean graph it manually without a calculator, or do you mean something else? Could calculus methods such as intervals where the graph increases or decreases be used? How about end behavior? Could/should you use the transformation of a standard graph using translations and expansions?
$endgroup$
– Rory Daulton
yesterday