What is the difference between shifting functions and shifting parabolas? The Next CEO of Stack OverflowExplaining Horizontal Shifting and ScalingMy solution is right and the book is wrong (parabolas) or did I misunderstand it?Finding the distance from a parabola (ballistic trajectory) to a point (for use in collision detection)Equation of normals at the end of variable chord of parabola $y^2-4y-2x=0$The Reason for different Forms of EquationsNon-symmetrical parabolaDoes the symmetry of a parabola in finding the maximum area of a rectangle under said parabola matter?Finding area of rectangle under a parabola asymmetrical with respect to the Y-axis: What did I do wrong?Co-ordinate Parabola Circle Contained in it; Difference in maximum and minimum possible radiusRelationship between parabolas and hyperbolasThe common tangent of two tilted parabolas
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What is the difference between shifting functions and shifting parabolas?
The Next CEO of Stack OverflowExplaining Horizontal Shifting and ScalingMy solution is right and the book is wrong (parabolas) or did I misunderstand it?Finding the distance from a parabola (ballistic trajectory) to a point (for use in collision detection)Equation of normals at the end of variable chord of parabola $y^2-4y-2x=0$The Reason for different Forms of EquationsNon-symmetrical parabolaDoes the symmetry of a parabola in finding the maximum area of a rectangle under said parabola matter?Finding area of rectangle under a parabola asymmetrical with respect to the Y-axis: What did I do wrong?Co-ordinate Parabola Circle Contained in it; Difference in maximum and minimum possible radiusRelationship between parabolas and hyperbolasThe common tangent of two tilted parabolas
$begingroup$
For shifting a parabola to the right, do we write a "$+$" sign or "$-$" sign in the equation? Is this the same way for shifting a function, as well?
Here is the equation of parabola just for the reference:
$y=(x-0)^2+c $
(Where $c$ is a variable; Y-intercept.)
(Correct me if I'm wrong.)
Another question, is there a "$-$" sign already there in an ideal parabola equation?
P.S. Apologies for the simplicity of the question.
conic-sections
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
$endgroup$
add a comment |
$begingroup$
For shifting a parabola to the right, do we write a "$+$" sign or "$-$" sign in the equation? Is this the same way for shifting a function, as well?
Here is the equation of parabola just for the reference:
$y=(x-0)^2+c $
(Where $c$ is a variable; Y-intercept.)
(Correct me if I'm wrong.)
Another question, is there a "$-$" sign already there in an ideal parabola equation?
P.S. Apologies for the simplicity of the question.
conic-sections
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
$endgroup$
$begingroup$
It's easy to see what happens if you just look at a simple example. Starting with $y=x^2$, going to $y=x^2+3$ shifts the graph up by three units; going instead to $y=(x+3)^2$ shifts the graph to the left by three units (if you don't see why, sketch the graph!).
$endgroup$
– Gerry Myerson
Mar 20 at 11:55
$begingroup$
Your question really isn't about parabolas. This may help: math.stackexchange.com/questions/133185/…
$endgroup$
– Ethan Bolker
Mar 20 at 12:08
add a comment |
$begingroup$
For shifting a parabola to the right, do we write a "$+$" sign or "$-$" sign in the equation? Is this the same way for shifting a function, as well?
Here is the equation of parabola just for the reference:
$y=(x-0)^2+c $
(Where $c$ is a variable; Y-intercept.)
(Correct me if I'm wrong.)
Another question, is there a "$-$" sign already there in an ideal parabola equation?
P.S. Apologies for the simplicity of the question.
conic-sections
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
$endgroup$
For shifting a parabola to the right, do we write a "$+$" sign or "$-$" sign in the equation? Is this the same way for shifting a function, as well?
Here is the equation of parabola just for the reference:
$y=(x-0)^2+c $
(Where $c$ is a variable; Y-intercept.)
(Correct me if I'm wrong.)
Another question, is there a "$-$" sign already there in an ideal parabola equation?
P.S. Apologies for the simplicity of the question.
conic-sections
conic-sections
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
edited Mar 20 at 11:27
Tanmay Patel
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
asked Mar 20 at 11:17
Tanmay PatelTanmay Patel
142
142
$begingroup$
It's easy to see what happens if you just look at a simple example. Starting with $y=x^2$, going to $y=x^2+3$ shifts the graph up by three units; going instead to $y=(x+3)^2$ shifts the graph to the left by three units (if you don't see why, sketch the graph!).
$endgroup$
– Gerry Myerson
Mar 20 at 11:55
$begingroup$
Your question really isn't about parabolas. This may help: math.stackexchange.com/questions/133185/…
$endgroup$
– Ethan Bolker
Mar 20 at 12:08
add a comment |
$begingroup$
It's easy to see what happens if you just look at a simple example. Starting with $y=x^2$, going to $y=x^2+3$ shifts the graph up by three units; going instead to $y=(x+3)^2$ shifts the graph to the left by three units (if you don't see why, sketch the graph!).
$endgroup$
– Gerry Myerson
Mar 20 at 11:55
$begingroup$
Your question really isn't about parabolas. This may help: math.stackexchange.com/questions/133185/…
$endgroup$
– Ethan Bolker
Mar 20 at 12:08
$begingroup$
It's easy to see what happens if you just look at a simple example. Starting with $y=x^2$, going to $y=x^2+3$ shifts the graph up by three units; going instead to $y=(x+3)^2$ shifts the graph to the left by three units (if you don't see why, sketch the graph!).
$endgroup$
– Gerry Myerson
Mar 20 at 11:55
$begingroup$
It's easy to see what happens if you just look at a simple example. Starting with $y=x^2$, going to $y=x^2+3$ shifts the graph up by three units; going instead to $y=(x+3)^2$ shifts the graph to the left by three units (if you don't see why, sketch the graph!).
$endgroup$
– Gerry Myerson
Mar 20 at 11:55
$begingroup$
Your question really isn't about parabolas. This may help: math.stackexchange.com/questions/133185/…
$endgroup$
– Ethan Bolker
Mar 20 at 12:08
$begingroup$
Your question really isn't about parabolas. This may help: math.stackexchange.com/questions/133185/…
$endgroup$
– Ethan Bolker
Mar 20 at 12:08
add a comment |
1 Answer
1
active
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$begingroup$
Let $f_0(x)$ be the function in hand with roots $r_i_i=1^i=n$ and let the roots of the shifted function $f_1(x)$ be $r_i'_i=1^n$. Clearly when we shift a function its roots must change in correspondence with the roots of the original function.
To the right: Since we're to shift to the right that means the corresponding new roots will be greater than the original roots i.e. $r_i'gt r_i$. So for a shift by $r_i'-r_i$ to the right the new function would be $f(x-(r_i'-r_i))$.
To the left: Since we're to shift to the left that means the corresponding new roots will be smaller than the original roots i.e. $r_i'lt r_i$. So for a shift by $r_i'-r_i$ to the left the new function would be $f(x+(r_i'-r_i))$.
Notice that Parabolas are functions described by Quadratic Equations, so these rules apply to them as well. You may also consider another approach by figuring out the vertex from the function.
Take the example of $f(x)=(x+5)^2$. Zeroes are $-5, -5$. Let's say you want to shift it by $5$ to the right. Then the function becomes $g(x)=f(x-5)=left((x-5)+5right)^2=x^2$ which is your required function.
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
$endgroup$
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
add a comment |
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$begingroup$
Let $f_0(x)$ be the function in hand with roots $r_i_i=1^i=n$ and let the roots of the shifted function $f_1(x)$ be $r_i'_i=1^n$. Clearly when we shift a function its roots must change in correspondence with the roots of the original function.
To the right: Since we're to shift to the right that means the corresponding new roots will be greater than the original roots i.e. $r_i'gt r_i$. So for a shift by $r_i'-r_i$ to the right the new function would be $f(x-(r_i'-r_i))$.
To the left: Since we're to shift to the left that means the corresponding new roots will be smaller than the original roots i.e. $r_i'lt r_i$. So for a shift by $r_i'-r_i$ to the left the new function would be $f(x+(r_i'-r_i))$.
Notice that Parabolas are functions described by Quadratic Equations, so these rules apply to them as well. You may also consider another approach by figuring out the vertex from the function.
Take the example of $f(x)=(x+5)^2$. Zeroes are $-5, -5$. Let's say you want to shift it by $5$ to the right. Then the function becomes $g(x)=f(x-5)=left((x-5)+5right)^2=x^2$ which is your required function.
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
$endgroup$
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
add a comment |
$begingroup$
Let $f_0(x)$ be the function in hand with roots $r_i_i=1^i=n$ and let the roots of the shifted function $f_1(x)$ be $r_i'_i=1^n$. Clearly when we shift a function its roots must change in correspondence with the roots of the original function.
To the right: Since we're to shift to the right that means the corresponding new roots will be greater than the original roots i.e. $r_i'gt r_i$. So for a shift by $r_i'-r_i$ to the right the new function would be $f(x-(r_i'-r_i))$.
To the left: Since we're to shift to the left that means the corresponding new roots will be smaller than the original roots i.e. $r_i'lt r_i$. So for a shift by $r_i'-r_i$ to the left the new function would be $f(x+(r_i'-r_i))$.
Notice that Parabolas are functions described by Quadratic Equations, so these rules apply to them as well. You may also consider another approach by figuring out the vertex from the function.
Take the example of $f(x)=(x+5)^2$. Zeroes are $-5, -5$. Let's say you want to shift it by $5$ to the right. Then the function becomes $g(x)=f(x-5)=left((x-5)+5right)^2=x^2$ which is your required function.
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
$endgroup$
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
add a comment |
$begingroup$
Let $f_0(x)$ be the function in hand with roots $r_i_i=1^i=n$ and let the roots of the shifted function $f_1(x)$ be $r_i'_i=1^n$. Clearly when we shift a function its roots must change in correspondence with the roots of the original function.
To the right: Since we're to shift to the right that means the corresponding new roots will be greater than the original roots i.e. $r_i'gt r_i$. So for a shift by $r_i'-r_i$ to the right the new function would be $f(x-(r_i'-r_i))$.
To the left: Since we're to shift to the left that means the corresponding new roots will be smaller than the original roots i.e. $r_i'lt r_i$. So for a shift by $r_i'-r_i$ to the left the new function would be $f(x+(r_i'-r_i))$.
Notice that Parabolas are functions described by Quadratic Equations, so these rules apply to them as well. You may also consider another approach by figuring out the vertex from the function.
Take the example of $f(x)=(x+5)^2$. Zeroes are $-5, -5$. Let's say you want to shift it by $5$ to the right. Then the function becomes $g(x)=f(x-5)=left((x-5)+5right)^2=x^2$ which is your required function.
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
$endgroup$
Let $f_0(x)$ be the function in hand with roots $r_i_i=1^i=n$ and let the roots of the shifted function $f_1(x)$ be $r_i'_i=1^n$. Clearly when we shift a function its roots must change in correspondence with the roots of the original function.
To the right: Since we're to shift to the right that means the corresponding new roots will be greater than the original roots i.e. $r_i'gt r_i$. So for a shift by $r_i'-r_i$ to the right the new function would be $f(x-(r_i'-r_i))$.
To the left: Since we're to shift to the left that means the corresponding new roots will be smaller than the original roots i.e. $r_i'lt r_i$. So for a shift by $r_i'-r_i$ to the left the new function would be $f(x+(r_i'-r_i))$.
Notice that Parabolas are functions described by Quadratic Equations, so these rules apply to them as well. You may also consider another approach by figuring out the vertex from the function.
Take the example of $f(x)=(x+5)^2$. Zeroes are $-5, -5$. Let's say you want to shift it by $5$ to the right. Then the function becomes $g(x)=f(x-5)=left((x-5)+5right)^2=x^2$ which is your required function.
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
edited Mar 20 at 12:14
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
answered Mar 20 at 12:05
Paras KhoslaParas Khosla
2,758423
2,758423
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
add a comment |
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Wait... hold on a sec... If we're going to the right, shouldn't we be subtracting inside the bracket, instead of adding?
$endgroup$
– Tanmay Patel
Mar 20 at 13:39
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
$begingroup$
Yes that is what I've done.
$endgroup$
– Paras Khosla
Mar 20 at 14:09
add a comment |
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$begingroup$
It's easy to see what happens if you just look at a simple example. Starting with $y=x^2$, going to $y=x^2+3$ shifts the graph up by three units; going instead to $y=(x+3)^2$ shifts the graph to the left by three units (if you don't see why, sketch the graph!).
$endgroup$
– Gerry Myerson
Mar 20 at 11:55
$begingroup$
Your question really isn't about parabolas. This may help: math.stackexchange.com/questions/133185/…
$endgroup$
– Ethan Bolker
Mar 20 at 12:08