Hello! Here's a question that I need help with. Thanks!
1) Construct a cubic equation that describes a graph that has no stationary points. Describe how you arrived at your equation.
2) Quadratic equations may describe graphs that do not intersect the x-axis at all. Is it possible to construct a cubic equation that describes a graph that does not intesect the x-axis at all? [My answer is no, but I don't really know how to explain it.] Why/Why not? State and explain an implication of your result using the Remainder-Factor Theorem.
Thank you in advance!
2) Your answer is correct.
From the general equation of a cubic it's clear that the range of a cubic is all real number numbers. In other words, . In order to go from -oo to +oo (or vice versa) the cubic has to cross the x-axis. The fancy theorem that formalises this is called the intermediate value theorem.
The implication is that all cubics have at least one real linear factor => at least one real root. Not sure why you'd bring in the remainder-factor theorem .....
Thank you very much for answering my questions! The document you provided is going to be really useful for me since we are going to be learning more about cubic functions.
If I am not wrong, perhaps the part of the question that says: " State and explain an implication of your result using the Remainder-Factor Theorem." requires some sort of conclusion that a cubic function can always be expressed/factorized to the following 3 forms?
[inspired by a part of the document you provided, so thank you!]
Since there will always be at least 1 root for a cubic equation,
y= (linear factor)(irreducible quadratic factor) OR
y= (linear factor)^2 (irreducible quadratic factor)
y= (linear factor)(linear factor)(linear factor)
Does that mean that once we know a root of the equation, we can always find the factor too? (It should be so, right?)
Some examples to illustrate:
1. y= (linear factor)(irreducible quadratic factor): One x-intercept.
y = (x + 2)(x^2 + x + 2) = x^3 + 3x^2 + 4x + 4. Has a NON-Stationary point of inflexion and NO stationary points.
y = (x - 1)(x^2 + x + 1) = x^3 - 1. Has a stationary point of inflexion - its only stationary point.
2. y= (linear factor)(linear factor)(linear factor): At least two x-intercepts.
y = (x - 1)(x + 1)(x + 2). Has two stationary points (max turning point and min turning point) and also a (non-stationary) point of inflection. Has three x-intercepts.
y = (x - 1)^2 (x + 1). Has two stationary points (max turning point and min turning point) and also a (non-stationary) point of inflection. The min turning point is also an x-intercept (at x = 1). Has two x-intercepts.
Okiedokie. Thank you for correcting my mistake! (I didn't even notice it!) >< Sorry to trouble you again, could you explain the difference between a stationary point of inflexion and a non-stationary point of inflexion?
There are two sorts of points of inflexions:
At a stationary point of inflexion the gradient of the tangent is zero (ie. f'(x) = 0).
At a NON-stationary point of inflexion the gradient of the curve is NOT zero.
Why is a point of inflexion spcial enough to get its own name:
At any point of inflexion, the concavity of the curve changes. The meaning of concavity boils down to whether the tangent is above or below the curve. Cocave down: the graph lies below its tangent. Concave up: the graph lies above its tangent. So at a point where the concavity of a curve changes (ie. at a point of inflexion), the tangent switches from being below to above (or vice versa) the graph.
Points of inflexion can be found by solving f''(x) = 0. But note that solutions to f''(x) = 0 are not always points of inflexion ..... Each solution must be tested.