Why should you sketch the graphs before solving a modulus equation?

The sketch shows how many solutions there are, and which piece of each graph, the original part or the reflected part, produces each one. Without it you cannot tell which equation to solve, and working algebraically alone can produce values that are not actually solutions.

How does solving a modulus inequality differ from solving the equation?

The graph work is identical: sketch both sides including the reflected parts, and find where they cross. Rather than stopping at those crossing points you then read off the regions between or beyond them, taking care over whether each endpoint is included.

Why might more than one cubic fit a given modulus graph?

Because the modulus hides which parts of the original curve were reflected, so the picture alone does not say what was below the axis. The original could have been either a positive or a negative cubic, and both may be consistent with what is shown.

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Equations, Inequalities & Graphs (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontModulus Functions
How do you sketch $y = |f (x)|$ from the graph of $y = f (x)$ ?

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How do you sketch $y = |f (x)|$ from the graph of $y = f (x)$ ?
Sketch $y = f (x)$ first, then reflect every part that lies below the $x$ -axis in the $x$ -axis.
Anything already on or above the axis stays exactly where it is, because those values are not negative.
The graph of $y = f (x)$ has a $y$ -intercept of $- 6$ . What is the $y$ -intercept of $y = |f (x)|$ ?
It is $6$ .
The reflection sends the point $(0, - 6)$ to $(0, 6)$ , because the modulus replaces each negative value by its positive equivalent.
True or False?
The graph of $y = |f (x)|$ is smooth where it meets the $x$ -axis.
False.
At those points there is a sharp V-shaped corner, where the original curve and its reflected image meet at an angle.
A smooth curve at the bottom would be a turning point, which is a different feature altogether.
What does the graph of $y = - |f (x)|$ look like?
It lies entirely on or below the $x$ -axis, since every modulus value is then made negative.
The corners where it meets the axis point upwards, giving an upside-down V rather than a V.
Why should you sketch the graphs before solving a modulus equation?
The sketch shows how many solutions there are, and which piece of each graph, the original part or the reflected part, produces each one.
Without it you cannot tell which equation to solve, and working algebraically alone can produce values that are not actually solutions.
$x = 1$ satisfies $x - 4 = 2 x - 5$ , but it is not a solution of $|x - 4| = 2 x - 5$ . Why not?
Substituting gives $|1 - 4| = 3$ on the left, but $2 (1) - 5 = - 3$ on the right, and $3 \neq - 3$ .
A modulus can never be negative, so any value making the other side negative has to be rejected.
This is why solutions must always be checked in the original equation.
How does solving a modulus inequality differ from solving the equation?
The graph work is identical: sketch both sides including the reflected parts, and find where they cross.
Rather than stopping at those crossing points you then read off the regions between or beyond them, taking care over whether each endpoint is included.
How does the sign of $a$ affect the graph of $y = a x^{3} + b x^{2} + c x + d$ ?
If $a > 0$ it is a positive cubic, coming up from the bottom left and leaving at the top right.
If $a < 0$ it is a negative cubic, running from the top left down to the bottom right.
What form does a cubic need to be in before you can sketch it?
Factorised, as a product of three linear factors, because that is what hands you the roots.
For example $2 x^{3} - 3 x^{2} - 11 x + 6$ is sketched from its factorised form $(2 x - 1) (x + 2) (x - 3)$ .
How do you find the $y$ -intercept of $y = (2 x - 1) (x - 3)^{2}$ ?
Substitute $x = 0$ and multiply the constants together:
$y = (- 1) (- 3)^{2} = - 9$
Every cubic has a $y$ -intercept, however many times it meets the $x$ -axis.
What does a repeated root look like on a cubic graph?
The curve touches the $x$ -axis there and turns back, rather than crossing it.
That touching point is also a turning point of the curve.
True or False?
If $f (x)$ is a cubic, the graph of $y = |f (x)|$ touches the $x$ -axis but never crosses it.
True.
A modulus is never negative, so the curve cannot pass below the axis: at each root it comes down to the axis and turns back.
There is always at least one such point, because every cubic crosses the $x$ -axis at least once.
You are deducing $f (x) = a (x - b) (x - c) (x - d)$ from a graph and you already have the roots. How do you find $a$ ?
Use the $y$ -intercept, which is the product $a \times (- b) \times (- c) \times (- d)$ .
Setting that equal to the intercept on the graph gives a single equation for $a$ , which is needed because many different cubics share the same roots.
Why might more than one cubic fit a given modulus graph?
Because the modulus hides which parts of the original curve were reflected, so the picture alone does not say what was below the axis.
The original could have been either a positive or a negative cubic, and both may be consistent with what is shown.
How do you solve a cubic inequality graphically?
Rearrange so that one side is zero, factorise the cubic, and sketch it from its roots.
Then read off the stretches of $x$ where the curve is on the required side of the axis, using the inequality sign to decide whether the roots themselves are included.
True or False?
You need the $y$ -intercept in order to solve a cubic inequality from a sketch.
False.
Only the roots are needed, because they are the points where the curve changes sign.
The $y$ -intercept says nothing about which stretches of the graph satisfy the inequality.
Solving $(3 x - 1) (x - 3)^{2} > 0$ gives $x > \frac{1}{3}$ together with $x \neq 3$ . Why is $x = 3$ excluded?
At $x = 3$ the repeated root makes the expression equal to zero, and the inequality is strict.
The curve only touches the axis there, so it is positive on both sides of $x = 3$ but not at that point itself.

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