On the 25th of every month, Tamesha pays her rent

How many points does the given equation have in common with the x-axis and where is the vertex in relation to the x-axis. y=x^2-

Fudgin [204]

Answer:

1. It has two points in common with the x-axis.

2. The vertex in relation to the x-axis is at $x=6$

Step-by-step explanation:

The points that the equation has in common with the x-axis are the points of intersection of the parabola with the x-axis.

To find them, substitute y=0 and solve for "x":

$y=x^2-12x+12\\0=x^2-12x+12$

Use the Quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\x=\frac{-(-12)\±\sqrt{(-12)^2-4(1)(12)}}{2(1)}\\\\x_1=10.89\\\\x_2=1.10$

It has two points in common with the x-axis.

To find the vertex in relation to the x-axis, use the formula:

$x=\frac{-b}{2a}$

Substituting values, you get:

$x=\frac{-(-12)}{2(1)}\\x=6$

3 0

3 years ago

What is a simple way to solve for the sum and difference of 2 cubes? For example, 27+64<img src="https://tex.z-dn.net/?f=x%5E3"

salantis [7]

Answer:

(3 + 4x)(9 - 12x + 16x²) = 0

(4m - 1)(16m² + 4m + 1) = 0

Step-by-step explanation:

Here we have to solve the sum of two cubes which is $27 + 64x^{3}$

Now, the equation is $27 + 64x^{3} = 0$

⇒ 3³ + (4x)³ = 0

⇒ (3 + 4x)[3² - 3(4x) + (4x)²] = 0

⇒ (3 + 4x)(9 - 12x + 16x²) = 0

So, (3 + 4x) = 0 or (9 - 12x + 16x²) = 0

Therefore, from the above two relation we can solve for x.

One root will be $- \frac{3}{4}$ and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.

Again, we have to solve the difference of two cubes which is $64m^{3} - 1$

Now, the equation is $64m^{3} - 1 = 0$

⇒ (4m)³ - 1³ = 0

⇒ (4m - 1)[(4m)² + 4m(1) + 1²] = 0

⇒ (4m - 1)(16m² + 4m + 1) = 0

So, (4m - 1) = 0 or (16m² + 4m + 1) = 0

Therefore, from the above two relation we can solve for m.

One root will be $\frac{1}{4}$ and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation. (Answer)