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The line y = ax + b is tangent to the graph

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Hi MHB,

I've come across a math problem lately and it seems so interesting to me but I don't understand the statement below, which caused me failed to think of a good method to solve it.

"The line is tangent to the graph at exactly two distinct points."

I understand that if we have a function, says, $y=(x^2-4)^2$, then $y=0$ is a tangent line to the curve at two distinct points, namely $(-2,0)$ and $(2,0)$.

Tangent to the curve.JPG

But in the problem as stated below, I honestly don't see how could a straight line can be a tangent to the given curve at two distinct points.

Problem:
The line $y=ax+b$ is tangent to the graph of $y=x^4-2x^3-9x^2+2x+8$ at exactly two distinct points. What is the value of $| a+b| $?

The only thing that I could think of to "force" the line $y=ax+b$ be the tangent to the curve is by drawing the green line that touches the curve at its extrema. This is a wrong tangent line, of course because the real tangent line at the extrema has zero slope...I think I am missing something (very important) here...

Any insight that anyone could give would be greatly appreciated.:)

Graph.JPG
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Re: The line y=ax+b is tangent to the graph

The only thing that I could think of to "force" the line $y=ax+b$ be the tangent to the curve is by drawing the green line that touches the curve at its extrema. This is a wrong tangent line, of course because the real tangent line at the extrema has zero slope...I think I am missing something (very important) here...
Hi anemone!

You can strike out the "of course", because a tangent line does not have to be horizontal.

In the extrema of a function, you do have a horizontal tangent, but that's not what this question is about.

The green line you have drawn is the right one.
Note that it does not touch the graph at its extrema, but slightly to the left where the slope is indeed angled down.

To solve the problem, you need to find 2 points with the same tangent slope $a$ (the derivative) and those 2 points also have to lie on both the curve and the line.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Re: The line y=ax+b is tangent to the graph

Hi anemone!

You can strike out the "of course", because a tangent line does not have to be horizontal.

In the extrema of a function, you do have a horizontal tangent, but that's not what this question is about.
:eek: I do realize that, I actually meant to say the tangent line at the extrema has zero slope...:)

The green line you have drawn is the right one.
Note that it does not touch the graph at its extrema, but slightly to the left where the slope is indeed angled down.
I don't get this, I like Serena, I'm sorry...(Sweating)

How much should we adjust the line in order to move it to the left and yet touches the curve at only two distinct points?

I have drawn it and I noticed that if I move the line slightly to the left, then line touches the curve only once...:(


Graph2.JPG
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Re: The line y=ax+b is tangent to the graph

Hi MHB,

I've come across a math problem lately and it seems so interesting to me but I don't understand the statement below, which caused me failed to think of a good method to solve it.

"The line is tangent to the graph at exactly two distinct points."

I understand that if we have a function, says, $y=(x^2-4)^2$, then $y=0$ is a tangent line to the curve at two distinct points, namely $(-2,0)$ and $(2,0)$.

View attachment 1475

But in the problem as stated below, I honestly don't see how could a straight line can be a tangent to the given curve at two distinct points.

Problem:
The line $y=ax+b$ is tangent to the graph of $y=x^4-2x^3-9x^2+2x+8$ at exactly two distinct points. What is the value of $| a+b| $?

The only thing that I could think of to "force" the line $y=ax+b$ be the tangent to the curve is by drawing the green line that touches the curve at its extrema. This is a wrong tangent line, of course because the real tangent line at the extrema has zero slope...I think I am missing something (very important) here...

Any insight that anyone could give would be greatly appreciated.:)

View attachment 1477
Setting $\displaystyle x_{1}$ and $\displaystyle x_{2}$ the abscissas of the tangents point You can write...

$\displaystyle y^{\ '} (x_{1}) = y^{\ '} (x_{2})$

$\displaystyle y(x_{2}) = y(x_{1}) + y^{\ '} (x_{1})\ (x_{2} - x_{1})\ (1)$

... and then solve (1) in the unknown quantities $\displaystyle x_{1}$ and $\displaystyle x_{2}$...


Kind regards


$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Re: The line y=ax+b is tangent to the graph

Problem:
The line $y=ax+b$ is tangent to the graph of $y=x^4-2x^3-9x^2+2x+8$ at exactly two distinct points. What is the value of $| a+b| $?
If the curve $y=x^4-2x^3-9x^2+2x+8$ touches the line $y=ax+b$ at two points then the difference $x^4-2x^3-9x^2+2x+8 - (ax+b) = x^4-2x^3-9x^2+(2-a)x+(8-b)$ will have two repeated roots, in other words it will be equal to $(x^2+px+q)^2 = x^4 + 2px^3 +(p^2+2q)x^2 + 2pqx + q^2$. Comparing coefficients of $x^3$ and $x^2$, you see that $p=-1$ and $q=-5$. Then compare the coefficients of $x$ and the constant term to see that $2-a=10$ and $8-b = 25$. Thus $a=-8$, $b=-17$ and the line is $y=-8x-17$.

Using that method, you get the additional information that the $x$-values at the points of tangency will be the roots of $x^2+px+q=0$, or $x^2-x-5=0$, namely $x = \frac12(1\pm\sqrt{21}).$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Re: The line y=ax+b is tangent to the graph

I don't get this, I like Serena, I'm sorry...

How much should we adjust the line in order to move it to the left and yet touches the curve at only two distinct points?

I have drawn it and I noticed that if I move the line slightly to the left, then line touches the curve only once...
I'm confused.
As far as I can tell you understand perfectly.

More specifically, your set of equations is:
\begin{cases}
y'(x_1)&=4x_1^3 - 6x_1^2-18x_1+2&=a \\
y'(x_2)&=4x_2^3 - 6x_2^2-18x_2+2&=a \\
y(x_1)&=x_1^4 - 2x_1^3-9x_1^2+2x_1+8&=ax_1 + b \\
y(x_2)&=x_2^4 - 2x_2^3-9x_2^2+2x_2+8&=ax_2 + b \\
q &= |a + b|
\end{cases}

The problem asks you to find $q$.



If you read off the y-coordinate of the green line in your graph at x=1, you can tell that $a+b \approx -25$.
So $q \approx 25$.

EDIT: From Opalg's response we can conclude that $q=25$.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,683
Re: The line y=ax+b is tangent to the graph

If the curve $y=x^4-2x^3-9x^2+2x+8$ touches the line $y=ax+b$ at two points then the difference $x^4-2x^3-9x^2+2x+8 - (ax+b) = x^4-2x^3-9x^2+(2-a)x+(8-b)$ will have two repeated roots, in other words it will be equal to $(x^2+px+q)^2 = x^4 + 2px^3 +(p^2+2q)x^2 + 2pqx + q^2$. Comparing coefficients of $x^3$ and $x^2$, you see that $p=-1$ and $q=-5$. Then compare the coefficients of $x$ and the constant term to see that $2-a=10$ and $8-b = 25$. Thus $a=-8$, $b=-17$ and the line is $y=-8x-17$.

Using that method, you get the additional information that the $x$-values at the points of tangency will be the roots of $x^2+px+q=0$, or $x^2-x-5=0$, namely $x = \frac12(1\pm\sqrt{21}).$
Thank you Opalg for the solution! :)

I'm confused.
As far as I can tell you understand perfectly.

More specifically, your set of equations is:
\begin{cases}
y'(x_1)&=4x_1^3 - 6x_1^2-18x_1+2&=a \\
y'(x_2)&=4x_2^3 - 6x_2^2-18x_2+2&=a \\
y(x_1)&=x_1^4 - 2x_1^3-9x_1^2+2x_1+8&=ax_1 + b \\
y(x_2)&=x_2^4 - 2x_2^3-9x_2^2+2x_2+8&=ax_2 + b \\
q &= |a + b|
\end{cases}

The problem asks you to find $q$.





If you read off the y-coordinate of the green line in your graph at x=1, you can tell that $a+b \approx -25$.
So $q \approx 25$.

EDIT: From Opalg's response we can conclude that $q=25$.
Hey I like Serena,

I want to thank you for your patience to guide me through the problem...it appeared that I was over thinking...

I understand it perfectly now, after reading the posts from Opalg and yours...

Thank you, I like Serena, for the help!:)