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S_A_V [24]
3 years ago
5

Rewrite equation in vertex form y=x^2+4x+2

Mathematics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer: y =(x+2)^2-2

Steps:

The general vertex form of a parabola is as follows:

y = a(x-x_v)^2+b

where xv is the x coordinate of the vertex, a is the coefficient determining how wide/narrow the parabola is and whether it is open-up (+) or open-down (-), and b is the bias (vertical shift.


Transforming an expression into the vertex form involves completing the square step:

y=x^2+4x+2\\y = x^2 + 2\cdot2 x+ 2 + 4 - 4= (x+2)^2-2\\y =(x+2)^2-2\\\implies x_v = -2,a=1, b=-2

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It’s 2 secant lines on a circle.

https://www.mathplanet.com/education/geometry/circles/advanced-information-about-circles

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3 years ago
harry owes the bank money.To replay his debt,he paid $150 back to the bank each month . After 10 months, his remaining debt was
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8 0
4 years ago
Let z1 = 2 − 2i and z2 = (1 − i) + √3(1 + i).
gtnhenbr [62]

Answer:

Step-by-step explanation:

z₁ = 2 − 2i

z₂ = (1 − i) + √3(1 + i) = (1 + √3) + (√3 - 1) i

a) We get the modulus of z₁ as follows

║z₁║ = √((2)²+(-2)²) = 2

now we find the argument

α = Arctan (-2/2) = Arctan (-1) = -45º  ⇒   α = 360º + (-45º) = 315º

b) z₁ = 2 Cis 315º

Although the complex number is in binomic or polar form, its representation must be the same, since the complex number is the same, only that it is expressed in two different forms. The modulus represents the distance from the origin to the point. The degree of  rotation is the angle from the x-axis. When the polar form is expanded, the result is  the rectangular form of a complex number.

c) If  z₀*z₁ = z₂  and   z₀ = a + b i

we have

(a + b i)*(2 − 2i) = (1 + √3) + (√3 - 1) i

⇒  2a + 2bi - 2ai - 2bi² = (1 + √3) + (√3 - 1) i

⇒  2a + 2bi - 2ai - 2b(-1) = (1 + √3) + (√3 - 1) i

⇒  2a + 2b + 2bi - 2ai = (1 + √3) + (√3 - 1) i

⇒ 2 (a + b) + 2 (b - a) i = (1 + √3) + (√3 - 1) i

Now we can apply

2 (a + b) = 1 + √3

2 (b - a) = √3 - 1

Solving the system we get

a = 1/2

b = √3 / 2

Finally

z₀ = (1/2) + (√3 / 2) i

d) ║z₀║ = √((1/2)²+(√3 / 2)²) = 1

α = Arctan ((√3 / 2)/(1/2)) = 60º

e) z₀ = Cis 60º

f) Since z₂ = z₀*z₁, then z₂ is the transformation of z₁ rotated counterclockwise by arg(w) which is 60º

8 0
3 years ago
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