Question

Y=x^2+4x+2 и y=2x^2-4x+9

Answer 1

Надо найти производную и приравнятьеё к нулю.  

Найти корни этого уравнения, так как так гда производная (в корнях )превращается в нуль там есть или экстремум или подозрение на экстремум.  

y=2x^4-9x^2+7  

y ' = 8x^3-18x 8x^3-18x =0 x^3-9/4 x =0  

Разложим на сомножители и легко увидим 3 корня кубического уравнения  

x * (x -3/2) (x+3/2) =0  

x= -3/2  

x=0  

x = +3/2  

Надо по правилам Анатолия Сергеевича найтии исследовать около каждого корня по 2 близлежащие точки (слева и справа) всего 6 точек. на расстоянии хотя бы  

0,25 от корня.  

Из кривой функции видно, что 3 экстремума явно выражены, сомнений нет.  

Answer 2

To find the points of intersection between the two functions $y = x^2 + 4x + 2$ and $y = 2x^2 - 4x + 9$, we need to set the two equations equal to each other and solve for $x$. Since both functions are equal to $y$, we have:

$x^2 + 4x + 2 = 2x^2 - 4x + 9$

Now, let's solve this equation step-by-step:

1. Subtract $2x^2$ from both sides of the equation to move all the terms to one side:

$2x^2 - x^2 + 4x + 2 - 2x^2 = 2x^2 - 2x^2 - 4x + 9$

This simplifies to:

$x^2 + 4x + 2 - 2x^2 = 9 - 4x$

2. Combine like terms:

$-x^2 + 4x + 2 = 9 - 4x$

3. Add $4x$ to both sides to get all the $x$-terms on one side:

$-x^2 + 4x + 2 + 4x = 9 - 4x + 4x$

This simplifies to:

$-x^2 + 8x + 2 = 9$

4. Move all the terms to one side by subtracting $9$ from both sides:

$-x^2 + 8x + 2 - 9 = 9 - 9$

This simplifies to:

$-x^2 + 8x - 7 = 0$

5. Now, we have a quadratic equation in standard form. To solve for $x$, we can use the quadratic formula:

The quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = -1$, $b = 8$, and $c = -7$.

$x = \frac{-8 \pm \sqrt{8^2 - 4(-1)(-7)}}{2(-1)}$

6. Calculate the discriminant:

$b^2 - 4ac = 8^2 - 4(-1)(-7) = 64 - 28 = 36$

7. Substitute the values into the quadratic formula:

$x = \frac{-8 \pm \sqrt{36}}{-2}$

8. Simplify:

$x = \frac{-8 \pm 6}{-2}$

Now, we have two possible solutions for $x$:

1. $x = \frac{-8 + 6}{-2} = \frac{-2}{-2} = 1$

2. $x = \frac{-8 - 6}{-2} = \frac{-14}{-2} = 7$

Now that we have the values of $x$, we can find the corresponding values of $y$ by substituting them back into either of the original equations. Let's use $y = x^2 + 4x + 2$ for this:

For $x = 1$:

$y = 1^2 + 4(1) + 2 = 1 + 4 + 2 = 7$

So, one point of intersection is $(1, 7)$.

For $x = 7$:

$y = 7^2 + 4(7) + 2 = 49 + 28 + 2 = 79$