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Archy [21]
2 years ago
10

How would you find the intercept for the equation y^2=2-3x-2x^2? I need the steps to understand

Mathematics
1 answer:
krek1111 [17]2 years ago
3 0

9514 1404 393

Answer:

  • x-intercepts: -2, +1/2
  • y-intercepts: ±√2

Step-by-step explanation:

You always find the intercepts by setting the known coordinate to its known value and solving for the value of the unknown coordinate.

An x-intercept is always found on the line y=0. Setting y=0 and solving for x, we find ...

  0 = 2 -3x -2x^2

  x^2 +3/2x -1 = 0 . . . . . . . . . . . . . . . . divide by -2

  (x2^2 +3/2x +(3/4)^2) -1 -(3/4)^2 = 0 . . . . . complete the square

  (x +3/4)^2 = 25/16 . . . . . add 25/16 and write as a square

  x +3/4 = ±5/4 . . . . . . . take the square root

  x = -3/4 ±5/4 = -8/4, +2/4

The x-intercepts are -2 and +1/2.

__

A y-intercept is always found on the line x=0. Setting x=0 and solving for y, we find ...

  y^2 = 2 -0 -0

  y = ±√2

The y-intercepts are -√2 and +√2.

_____

A graphing calculator can give you a good idea of what the intercepts are.

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Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )
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QUESTION 1

The given inequality is

|m|-2>0

We group like terms to get,

|m|>2


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m or m>2.

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(-\infty,-2)\cup (2,+\infty).


QUESTION 2

|x-4|-3\:>\:5.

We group like terms to get,


|x-4|\:>\:5+3.


|x-4|\:>\:8

We split the absolute value sign to get,

-(x-4)\:>\:8 or x-4\:>\:8


This implies that,


x-4\: or x-4\:>\:8


x\: or x\:>\:8+4


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We can write this interval notation to get,

(-\infty,-4)\cup (12,+\infty).


QUESTION 3

The given inequality is

|6+9x|\leq 24


We split the absolute value sign to obtain,

-(6+9x)\leq 24 or (6+9x)\leq 24


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6+9x\ge -24 and 6+9x\leq 24


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-\frac{10}{3}\leq x\leq2

We write this in interval form  to get,

[-\frac{10}{3},2]


QUESTION 4

The given inequality is

|1-5a|>29

We split the absolute value sign to get,

-(1-5a)>29 or 1-5a>29

This simplifies to,

1-5a\: or 1-5a\:>\:29


This implies that,

-5a\: or -5a\:>\:29-1


-5a\: or -5a\:>\:28


a\:>\:6 or a\:

We write this in interval notation to get,

(-\infty,-\frac{28}{5})\cup (6,+\infty)















7 0
3 years ago
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