final pratice exam help please no calculator

Question

Answer 1

1)

Given limit function is

= 2+1

= 3

Solution : Option (A) is Correct Answer.

Answer 2

2)

Given limit function is

As x is approaches to zero. x ≥ 0 condition is satisfied.

So substitute f(x)  = x - 1

= -1

Solution : Option (C) is Correct Answer.

 

Comments

How do you know x is greater but not less than 0? How do you know that x is approcaching zero from the high but not the low? How you determine that x was greater?

---

Your answer is wrong btw. I have the answer key and the answer is D

---

2)

In such cases we have to check for the condition first.

If above condition satisfies then only limit exists,Otherwise limit does not exists.

Both limits are not equal.So condition is not satisfied.

Solution : Option (d) is Correct Answer.

---

Why do you choose 0+ and 0-? is it a rule that you must use 0?

---

Yes that is the rule.

For a limit to exist at a particular value 'a',

the left handed limit a⁺ must be equal to right handed limit a^-.

Answer 3

3)

y = ( 2 + x ) / (1 - x )

y = ( x + 2 ) / ( - x + 1 )

Coefficient of Highest degree term of the Numerator polynomial x + 2 is 1.

Coefficient of Highest degree term of the Denominator polynomial - x + 1 is -1

Formula of Horizontal asymptote :

y = ( Coefficient of Highest degree term of the Numerator polynomial ) / ( Coefficient of Highest degree term of the Denominator polynomial )

y = (1) / (-1 )

y = -1

Solution : Option (e) is Correct Answer.

Comments

What? How did x + 2 become 1 and -x + 1 become -1? I don't get it.

---

The numerator is x + 2

Here the highest degree term in x + 2 is x.

Coefficient means the constant placed before the variable.

x can be written as (1)x

Therefore the coefficient of x in x + 2 is 1.

Similarly, the denominator is -x + 1

Here the highest degree term in -x + 1 is -x.

-x can be written as (-1)x

Therefore the coefficient of x in -x + 1 is -1.

Horizontal asymptote:

y = (Numerator's leading Coefficient) / (Denominator leading Coefficient)

y = 1 / (-1)

y = -1

Answer 4

4)

Given point is ( - 6 , 4 ).

x - Intercept is c = 3. So line is passing through point ( 3 , 0 ).

Slope m = ( y_{2 -} y₁ ) / (x_{2 -} x₁) = ( 4 – 0 ) / ( - 6 – 3 ) = 4 / (-9) = - 4/9

 

Slope (m) point ( x₁ , y₁ ) form of straight line is

( y - y₁ ) = m ( x – x₁ )

Substitute ( x₁ , y₁ ) = ( 3 , 0 ) and m - 4/9 in the above equation.

( y - 0 ) = (- 4/9 ) ( x – 3 )

y = (- 4/9 ) x - 3 (- 4/9)

y = (- 4/9 ) x + 4/3.

Solution : Option (e) is Correct Answer