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kozerog [31]
3 years ago
10

Complete the square to rewrite x^2+y^2+2x+6y-6=0 in graphing form

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

(x + 1)² + (y + 3)² = 16

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + y² + 2x + 6y - 6 = 0

Collect the x and y terms together and add 6 to both sides

x² + 2x + y² + 6y = 6

To complete the square

add ( half the coefficient of the x/ y terms )² to both sides

x² + 2(1)x + 1 + y² + 2(3)y + 9 = 6 + 1 + 9

(x + 1)² + (y + 3)² = 16

with centre = (- 1, - 3) and r = \sqrt{16} = 4

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Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated
Illusion [34]

Answer:

x=-2,0,3

Step-by-step explanation:

We have been given a function f(x)=15x^3-15x^2-90x. We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

15x^3-15x^2-90x=0

Now, we will factor our equation. We can see that all terms of our equation a common factor that is 15x.

Upon factoring out 15x, we will get:

15x(x^2-x-6)=0

Now, we will split the middle term of our equation into parts, whose sum is -1 and whose product is -6. We know such two numbers are -3\text{ and }2.

15x(x^2-3x+2x-6)=0

15x((x^2-3x)+(2x-6))=0

15x(x(x-3)+2(x-3))=0

15x(x-3)(x+2)=0

Now, we will use zero product property to find the zeros of our given function.

15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0

15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0

\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0

x=0\text{ (or) }x=3\text{ (or) }x=-2

Therefore, the zeros of our given function are x=-2,0,3.

7 0
3 years ago
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