# Minima Maxima: y=ax³+bx²+cx+d

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Francis June E.... Minima Maxima: y=ax³+bx²+cx+d

What is the condition that the cubic y=ax³+bx²+cx+d shall have two Extremes?

Salamat po.

Jhun Vert Differentiate the given cubic equation and equate to zero. Solve the roots of the resulting quadratic equation. To have two extremes, the roots of the quadratic equation must be real. What makes the roots real is when the discriminant of the quadratic equation greater than zero.

If you have further question, don't hesitate to ask.

Francis June E.... D ko po makuha yung roots sir

Jhun Vert $y = ax^3 + bx^2 + cx + d$

$y' = 3ax^2 + 2bx + c = 0$

The roots of the quadratic equation are
$x = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

$x = \dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a}$

To have two extremes
$4b^2 - 12ac \gt 0$

Francis June E.... Yan na po value ni x?

Francis June E.... Paano po makuha c 4b²-12ac?

Jhun Vert Substitute:
A = 3a
B = 2b
C = c

Francis June E.... When i substitute x to y", answer nya po is 2√b²-3ac and -2√b²-3ac?

Pero yung answer daw po is only b²-3ac.
How po?

Jhun Vert You don't need to do that. Divide 4b2 - 12ac > 0 both sides by 4 and you will get the answer.

Francis June E.... Pero paano po nakuha yung 4b²-12ac?

Francis June E.... When i substitute A,B and C

Paano po naging 4b²-12ac nalang po?

Francis June E.... D ko parin po makuha kung paano naging b²-3ac nalang

Jhun Vert The problem ask a condition for the cubic to have two extremes. You give "4b2 - 12ac > 0" as your condition to satisfy the problem.

From $x = \dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a}$

• If $4b^2 - 12ac \lt 0$, x is imaginary. (no extreme)
• If $4b^2 - 12ac = 0$, you only have one value of x. (one extreme)
• If $4b^2 - 12ac \gt 0$, there are two values of x. (two extremes)

That is why you give $4b^2 - 12ac \gt 0$ as you answer to satisfy the condition of the problem.

Francis June E.... Okay na po. Salamat po .

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