# Math Help - Graphs of basic functions & Remainder-Factor theorem

1. ## Graphs of basic functions & Remainder-Factor theorem

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.

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Thank you very much!

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

3. I see! Thanks for explaining!

4. 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.

5. 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.

6. 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?

7. 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.

8. 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.

9. 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.

10. 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!

11. 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.