Other Points Where Tangent Line Intersects Graph

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Q: For each a ∈ $R$ find any other points at which the tangent line ($y = 3a^2 -48$) intersects the original graph ($x^3 - 48x + 2$). Hint: $f(x) − f(a)$ is divisible by $x − a$
Does this just mean whenever $x-a not= 0$?
calculus derivatives tangent-line
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Q: For each a ∈ $R$ find any other points at which the tangent line ($y = 3a^2 -48$) intersects the original graph ($x^3 - 48x + 2$). Hint: $f(x) − f(a)$ is divisible by $x − a$
Does this just mean whenever $x-a not= 0$?
calculus derivatives tangent-line
New contributor
Emma Pascoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Q: For each a ∈ $R$ find any other points at which the tangent line ($y = 3a^2 -48$) intersects the original graph ($x^3 - 48x + 2$). Hint: $f(x) − f(a)$ is divisible by $x − a$
Does this just mean whenever $x-a not= 0$?
calculus derivatives tangent-line
New contributor
Emma Pascoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Q: For each a ∈ $R$ find any other points at which the tangent line ($y = 3a^2 -48$) intersects the original graph ($x^3 - 48x + 2$). Hint: $f(x) − f(a)$ is divisible by $x − a$
Does this just mean whenever $x-a not= 0$?
calculus derivatives tangent-line
calculus derivatives tangent-line
New contributor
Emma Pascoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Emma Pascoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Emma Pascoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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First of all, note that you are given a $y=k$ type line where $k$ is a constant.
For this line, the slope comes out to be $0$. Time to find out at which point the slope is $0$:
$$ frac{mathrm{d} }{mathrm{d} x}left ( x^{3}-48x+2 right ) = 3x^{2}-48 $$
Given that it is equal to $0$, there are two values of $x$, that is $x=pm 4$. The two points are $ (4, -126)$ and $(-4,130)$, so the two lines are now $y=130$ and $y=-126$.
In order to find the where the graph cuts again we just plug in these value in $f(x)=126 text{ or } 130$. These give the value $x=8$ for $130$ and $x=-8$ for $-126$.
Now note that $y=3a^2-48=-126$ cannot be used because of the fact that $a$ is not real here. So the final answer is $(8, 130)$.
As from the hint, I suggest you to read Remainder Factor Theorem. And in order to solve assume a new function $g(x)=f(x)-f(a)$, and apply remainder theorem.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
First of all, note that you are given a $y=k$ type line where $k$ is a constant.
For this line, the slope comes out to be $0$. Time to find out at which point the slope is $0$:
$$ frac{mathrm{d} }{mathrm{d} x}left ( x^{3}-48x+2 right ) = 3x^{2}-48 $$
Given that it is equal to $0$, there are two values of $x$, that is $x=pm 4$. The two points are $ (4, -126)$ and $(-4,130)$, so the two lines are now $y=130$ and $y=-126$.
In order to find the where the graph cuts again we just plug in these value in $f(x)=126 text{ or } 130$. These give the value $x=8$ for $130$ and $x=-8$ for $-126$.
Now note that $y=3a^2-48=-126$ cannot be used because of the fact that $a$ is not real here. So the final answer is $(8, 130)$.
As from the hint, I suggest you to read Remainder Factor Theorem. And in order to solve assume a new function $g(x)=f(x)-f(a)$, and apply remainder theorem.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
add a comment |
up vote
0
down vote
First of all, note that you are given a $y=k$ type line where $k$ is a constant.
For this line, the slope comes out to be $0$. Time to find out at which point the slope is $0$:
$$ frac{mathrm{d} }{mathrm{d} x}left ( x^{3}-48x+2 right ) = 3x^{2}-48 $$
Given that it is equal to $0$, there are two values of $x$, that is $x=pm 4$. The two points are $ (4, -126)$ and $(-4,130)$, so the two lines are now $y=130$ and $y=-126$.
In order to find the where the graph cuts again we just plug in these value in $f(x)=126 text{ or } 130$. These give the value $x=8$ for $130$ and $x=-8$ for $-126$.
Now note that $y=3a^2-48=-126$ cannot be used because of the fact that $a$ is not real here. So the final answer is $(8, 130)$.
As from the hint, I suggest you to read Remainder Factor Theorem. And in order to solve assume a new function $g(x)=f(x)-f(a)$, and apply remainder theorem.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
First of all, note that you are given a $y=k$ type line where $k$ is a constant.
For this line, the slope comes out to be $0$. Time to find out at which point the slope is $0$:
$$ frac{mathrm{d} }{mathrm{d} x}left ( x^{3}-48x+2 right ) = 3x^{2}-48 $$
Given that it is equal to $0$, there are two values of $x$, that is $x=pm 4$. The two points are $ (4, -126)$ and $(-4,130)$, so the two lines are now $y=130$ and $y=-126$.
In order to find the where the graph cuts again we just plug in these value in $f(x)=126 text{ or } 130$. These give the value $x=8$ for $130$ and $x=-8$ for $-126$.
Now note that $y=3a^2-48=-126$ cannot be used because of the fact that $a$ is not real here. So the final answer is $(8, 130)$.
As from the hint, I suggest you to read Remainder Factor Theorem. And in order to solve assume a new function $g(x)=f(x)-f(a)$, and apply remainder theorem.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First of all, note that you are given a $y=k$ type line where $k$ is a constant.
For this line, the slope comes out to be $0$. Time to find out at which point the slope is $0$:
$$ frac{mathrm{d} }{mathrm{d} x}left ( x^{3}-48x+2 right ) = 3x^{2}-48 $$
Given that it is equal to $0$, there are two values of $x$, that is $x=pm 4$. The two points are $ (4, -126)$ and $(-4,130)$, so the two lines are now $y=130$ and $y=-126$.
In order to find the where the graph cuts again we just plug in these value in $f(x)=126 text{ or } 130$. These give the value $x=8$ for $130$ and $x=-8$ for $-126$.
Now note that $y=3a^2-48=-126$ cannot be used because of the fact that $a$ is not real here. So the final answer is $(8, 130)$.
As from the hint, I suggest you to read Remainder Factor Theorem. And in order to solve assume a new function $g(x)=f(x)-f(a)$, and apply remainder theorem.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday


Lakshya Sinha
364
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
add a comment |
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
But how does the remainder theorem play into finding points?
– Emma Pascoe
yesterday
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
I wrote this answer in night, so I have now edited a few things. The remainder theorem gives us that g(x) is divisible by x-4 so the cubic can now be easily reduced to a quadratic, which can be easily solved
– Lakshya Sinha
17 hours ago
add a comment |
Emma Pascoe is a new contributor. Be nice, and check out our Code of Conduct.
Emma Pascoe is a new contributor. Be nice, and check out our Code of Conduct.
Emma Pascoe is a new contributor. Be nice, and check out our Code of Conduct.
Emma Pascoe is a new contributor. Be nice, and check out our Code of Conduct.
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