Graphs of basic functions & Remainder-Factor theorem

**Tangera** · January 22nd 2008, 05:53 AM

Hello! I have a few questions regarding the graphs of basic functions (eg. y = x^3). Please provide explanations to the answers if possible! Thank You very much!

Q1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

Q2) It is not possible to construct a cubic equation that describes a graph that does not intersect with the x-axis. Why not? State and explain using the Remainder and Factor Theorem.

Q3) Given that the 2 quadratic equations:
x^2 -nx -36 = 0
n + 6x - x^2 = 0
have a common root x = a for certain values of the real constant n, show that a^3 - 7(a^2) + 36 = 0

Hence find the possible values of a, and the corresponding values of n.

---
Thank you very much!

**Soroban** · January 22nd 2008, 09:12 AM

Hello, Tangera!

3) Given that the two quadratic equations: . $\begin{array}{ccc}x^2 -nx -36 & = & 0 \\ x^2-6x-n &=&0\end{array}$

have a common root $x = a$ for certain values of the real constant $n$ ,

(a) Show that: . $a^3 - 7a^2 + 36 \:= \:0$

Since $a$ is a root of both equations, we have:

. . $\begin{array}{cccccccc}a^2-an-36 &=&0 & \Longrightarrow & n & = &\dfrac{a^2-36}{a} & {\color{blue}[1]} \\ \\ a^2-6a-n &=&0 & \Longrightarrow & n &=&a^2-6a & {\color{blue}[2]} \end{array}$

Equate [1] and [2]: . $\frac{a^2-36}{a} \:=\:a^2-6a\quad\Longrightarrow\quad\boxed{ a^3 - 7a^2 + 36 \:=\:0}$

$\text{(b) Hence, find the possible values of }a\text{ and the corresponding values of }n.$

The cubic factors: . $(a-3)(a-6)(a+2) \:=\:0$

. . and has roots: . $a \;=\;3,\:6,\:-2$

Substitute into [2]: . $n \;=\;-9,\:0,\:16$

Therefore: . $\boxed{(a,\,n) \;=\;(3,\,\text{-}9),\;(6,\,0),\;(\text{-}2,\,16)}$

**Tangera** · January 23rd 2008, 02:22 AM

I see! Thanks for explaining!

**stonetoad** · January 26th 2008, 05:49 AM

For your first question,

For a graph to have a stationary point, dy/dx must be 0 at some point.
For a cubic graph, dy/dx is always a quadratic expression.
When can't a quadratic expression = 0? When the discriminant is < 0.
So, just find yourself a b^2-4ac that's < 0, then from there construct a quadratic expression. Then just integrate it for the cubic equation.

Hope I helped.

**angel.white** · January 26th 2008, 06:19 AM

Originally Posted by stonetoad

For your first question,

For a graph to have a stationary point, dy/dx must be 0 at some point.
For a cubic graph, dy/dx is always a quadratic expression.
When can't a quadratic expression = 0? When the discriminant is < 0.
So, just find yourself a b^2-4ac that's < 0, then from there construct a quadratic expression. Then just integrate it for the cubic equation.

Hope I helped.

Is that even possible? If the quadratic is the result of a derivative, then c = 0 always. So you are trying to find b^2 < 0. Only an imaginary number could fulfill this requirement.

**stonetoad** · January 26th 2008, 06:36 AM

Originally Posted by angel.white

Is that even possible? If the quadratic is the result of a derivative, then c = 0 always. So you are trying to find b^2 < 0. Only an imaginary number could fulfill this requirement.

Yes, it's possible. For example, let's say the discriminant is (2^2)-4(3)(3). So the quadratic expression that results in this discriminant is x^3 + 2x +3. Then you integrate it to get the cubic expression x^3 + x^2 + 3x. It all makes sense, doesn't it?

**angel.white** · January 26th 2008, 07:12 AM

Originally Posted by stonetoad

Yes, it's possible. For example, let's say the discriminant is (2^2)-4(3)(3). So the quadratic expression that results in this discriminant is x^3 + 2x +3. Then you integrate it to get the cubic expression x^3 + x^2 + 3x. It all makes sense, doesn't it?

Oh, gotcha.

**mr fantastic** · January 26th 2008, 12:08 PM

Originally Posted by Tangera

Hello! I have a few questions regarding the graphs of basic functions (eg. y = x^3). Please provide explanations to the answers if possible! Thank You very much!

Q1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.

Q2) It is not possible to construct a cubic equation that describes a graph that does not intersect with the x-axis. Why not? State and explain using the Remainder and Factor Theorem.

Q3) Given that the 2 quadratic equations:
x^2 -nx -36 = 0
n + 6x - x^2 = 0
have a common root x = a for certain values of the real constant n, show that a^3 - 7(a^2) + 36 = 0

Hence find the possible values of a, and the corresponding values of n.

---
Thank you very much!

So to summarise:

Questions 1 and 2 have been addressed here. Soroban has addressed question 3.

**mr fantastic** · January 26th 2008, 12:17 PM

Originally Posted by angel.white

Is that even possible? If the quadratic is the result of a derivative, then c = 0 always. So you are trying to find b^2 < 0. Only an imaginary number could fulfill this requirement.

All the theory (in the form of questions and solutions) is given in the attachment found here.

**Tangera** · January 26th 2008, 05:49 PM

Hello! Many thanks to everyone who pondered over the questions and helped answer them! I've learnt quite a bit from your answers and questions. Thank you very much!

**mr fantastic** · January 26th 2008, 07:10 PM

Originally Posted by Tangera

Hello! Many thanks to everyone who pondered over the questions and helped answer them! I've learnt quite a bit from your answers and questions. Thank you very much!

And on behalf of the forum, thankyou for your lovely manners.

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