Hello everyone, today I will show you how to calculate the area of a triangle created by the three extreme points of the function.

Usually, the problem shows you the function and then asks you to calculate area of trianglec, so your first job is to find the three extremes.

When you find the three extreme points, it means you have gone half of the way, the rest you just need to apply one of the ways below.

Okay, now we will get to the main content right now!

**Way #1. Based on the determinant of square matrix of order 2 x 2**

This way uses knowledge of Advanced Algebra, mainly knowledge of Linear Algebra.

**Step 1.** Find the three extreme points of the graph of the function f(x)

Suppose we find that $A=(x_a; y_a), B=(x_b; y_b), C=(x_c; y_c)$ are the three extreme points of the given function graph.

**Step 2.** Calculate vector $\overrightarrow{AB}=(x_b-x_a; y_b-y_a), \overrightarrow{AC}=(x_c-x_a; y_c-y_a)$

For convenience of presentation, here I will put $\overrightarrow{AB}=(a; b), \overrightarrow{AC}=(c; d)$

**Step 3.** Evaluate the expression $\left|\begin{array}{cc}a&b\\c&d\end{array}\right|=ad-cb$

**Step 4.** The value of the expression $\frac{|ad-cb|}{2}$ is the area of triangle ABC

**Example 1.** Calculate the area of the triangle formed by the three extreme points of the graph of the function $f(x)=3x^4-40x^3+186x^2-360x$

**The answer:**

It is easy to see, the three extreme points of the graph of the function $f(x)=3x^4-40x^3+186x^2-360x$ are $A=(2; -248), B=(3; -243), C=(5; -275)$

The vectors $\overrightarrow{AB}, \overrightarrow{AC}$ are $(1; 5), (3; -27)$ respectively

$\left|\begin{array}{cc}1&5\\3&-27\end{array}\right|=1(-27)-5.3=-42$

$S_{ABC}=\frac{|-42|}{2}=21$

So the area of triangle ABC is 21 units

**Method #2. Using Heron’s Formula**

In this way we will move from classical Calculus to Elementary Geometry thanks to the formula for calculating the distance between two points.

Assuming the point $A=(x_a; y_a), B=(x_b; y_b)$ then the distance between two points A, B will be calculated by the formula $\sqrt{(x_b-x_a)^2+(y_b -y_a)^2}$

**2.1. Regarding the solution steps as follows:**

**Step 1.** Calculate the lengths of sides AB, BC, CA

**Step 2.** Apply Heroon’s formula

For convenience of presentation, here I will put $AB=a, BC=b, CA=c, p=\frac{a+b+c}{2}$

Then the area of triangle ABC will be calculated by the formula $\sqrt{p(pa)(pb)(pc)}$

**Example 2.** Calculate the area of the triangle formed by the three extreme points of the graph of the function $f(x)=3x^4-40x^3+186x^2-360x$

**The answer:**

It is easy to see that the three extreme points of the graph of the function $f(x)=3x^4-40x^3+186x^2-360x$ are $A=(2; -248), B=(3; – 243), C=(5; -275)$

The lengths of the three sides AB, BC, CA are $\sqrt{26}, 2\sqrt{257}, 3\sqrt{82}$ respectively

Half circumference $\frac{\sqrt{26}+2\sqrt{257}+3\sqrt{82}}{2}$

So the area of triangle ABC is 21 units

**2.2. Simplify complex expressions with Casio calculator**

As you can see, the above expression is quite complicated, not to mention that you collapse “by hand”, even if you reduce it with the CASIO calculator by directly entering the expression, it still takes a lot of time. and prone to errors

The solution is to assign $\sqrt{26}, 2\sqrt{257}, 3\sqrt{82}, \frac{\sqrt{26}+2\sqrt{257}+3\sqrt{82}} {2}$ into the memory variables A, B, C, D, respectively, and then calculate the value of the expression $\sqrt{D(DA)(DB)(DC)}$

**How to assign a value to a memory variable as follows:**

Enter value => press key **STO** => **press the memory key.**

**As follows:**

Press the keys in turn…

**Implementation steps:**

**Step 1.** Assign values to memory variables.

**Step 2.** Enter the expression:

**Step 3.** Press `=`

**Way #3. Use special formula**

**Way 1** and **Way 2** can be applied to any function, as long as the function has three non-collinear extreme points.

And this way 3 is only applicable to the function $f(x)=ax^4+bx^2+c$ with the condition `a`

and `b`

against you guys.

At that time, the area of the triangle created by the three extreme points of the graph of the function $f(x)=ax^4+bx^2+c$ will be calculated by the formula $\sqrt{-\frac{b^5}{32a^3}}$

**Example 3.** Calculate the area of the triangle formed by the three extreme points of the graph of the function $f(x)=3x^4-7x^2$

**The answer:**

The area of the triangle formed by the three extreme points of the graph of the function $f(x)=3x^4-7x^2$ will be equal to $\sqrt{-\frac{(-7)^5}{32.( 3)^3}}=\frac{49\sqrt{42}}{72}$

So the area of the triangle to find is $\frac{49\sqrt{42}}{72} \approx 4.41$ unit

**#4. Epilogue**

Okay, here are 3 ways **calculate the area of the triangle formed by the three extreme points **that I want to share with you in this article. In most cases you should prefer using **Way 1** to solve the exercise.

**Way 3** although very fast, the applicability is quite limited, only if the given function is a quadratic function, **a** and **b** opposite sign can be applied.

In addition, to calculate the area of the triangle created by the three extreme points you should also review directional product, formula for area of triangle…

Hope this article will be useful to you. Goodbye and see you in the next posts!

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