Need help w/ 2 MATH PROBLEMS?

Question

1. Find f(0) for
f(x) =
x^2-3x +2
-------------
x+1

A. 4
B. 7
C. 2
D. 3

2. solve for f(x)= 0
f(x)=
x^2-3x +2
-------------
x +1

A. {1,2}
B. {-1,3}
C. {1,-1}
D. {1,-4}

Answer 1

1. C
2. A

Answer 2

For the first question, the "x" is just a placeholder. One could think of it as a box that needs filling. In this case, we fill it with the number zero. Since f(x) = (x^2 - 3x + 2)/(x+1), we have:

f(0) = (0^2 + 3(0) + 2)/(0+1) = 2/1 = 2

(This tells you what the function's y-value is at its y-intercept, by the way. In other words, the function's graph intercepts the y-axis at the point (0, 2).)

For the second question, we're using the same function - only this time we're trying to find out when its OUTPUT is zero. (In the first problem, we were just looking for what happens when its INPUT was zero.) So, we set the function equal to zero:

(x^2 - 3x + 2)/(x+1) = 0

and we solve for x. We can multiply both sides by x+1 and cancel out the denominator so long as we assume that x + 1 is not zero. That is, x cannot be -1. (We knew this already, actually, because -1 isn't in the domain of the function. Otherwise we'd be dividing by zero!) Therefore we just need to solve this equation:

x^2 - 3x + 2 = 0*(x+1)

But since zero times anything is zero, we really need to solve this quadratic equation:

x^2 - 3x + 2 = 0

Conveniently, this factors:

(x - 2)(x - 1) = 0

Therefore, this is true exactly when x = 2 or x = 1. Since neither of these values is -1 (which we said earlier can't be a solution due to the evils of division by zero), both are correct.

Graphically, this says that the graph of the function f touches the x-axis at the x-values of 1 and 2. In other words, its x-intercepts are (1,0) and (2,0).

Answer 3

for problem 1.) You are asked to find f(0). Take this problem into 3 smaller parts. x^2 -3x +2. Follow your mathematical rules and substitute your numbers in for X. (0)^2 -3(0) +2. Tackle (0)^2 first. (0)^2 is pretty much the same as 0*0 and that for certain is equal to 0. Next is -3(0). Any number multiplied by zero is always zero so this is also 0. Last number has no variable and remains at 2. So we have 0-0+2=2 So your numerator is 2. Now for the denominator. (X+1) can be substituted with (0+1) or just 1. so you have 2 for the numerator and 1 for the denominator or 2/1. Any number divided by one is unchanged so the answer will be 2 of the answer C. Try this and see if you can answer problem 2.

Answer 4

On the first problem all you have to do is replace the X term with 0. That is 0^2- 3(0)+2 / 0+1 [you do the math]

On the second problem you have to set your function equal to 0 and then you have an equation:
x^2-3x +2 / x + 1 = 0

which for this case it hapens to be the same as

x^2-3x +2 = 0

from there you can use quadratic formula or factorization to find the zeros.

Answer 5

Ans:1.C
Explanation:
f(0)=(0^2-3*0+2)/(0+1)
=2/1
=2
Ans:2 .A
Explanation:
f(x)=0
x^2-3x+2=0
(x-1)(x-2)=0
x=1 or x=2
Therefore the solution set is {1,2}

Answer 6

(x^2-3x+2)/x+1 =f(x)
(0 -0 +2)/(0 +1)=f(0) = 2 The answer is C

(x^2-3x+2)/(x+1) = 0
x^2 -3x +2 =0
(x-1)(x-2) =0
x=1 and x=2 the answer is A.

Answer 7

1 F(0) = (0^2 - 3*0 + 2) / ( 0 + 1 ) = 2/ 1 = 2

x^2 - 3x + 2 / (X+1) = 0

x^2 - 3x + 2 = 0

(X -2) * (x-1) = 0

X = 2 or 1

Answer 8

Question 1
f(0) = 0 - 0 + 2 = 2
ANSWER C

Question 2
x² - 3x + 2 = (x - 2).(x - 1) = 0
x = 1, x = 2
ANSWER A

Answer 9

1.f(x)=x^2-3x+2
f(0)=0^2 -3(0)+2
f(0) =2
Answer=C

2.f(x)=x^2-3x+2
f(x)=0
x^2-3x+2=0
x^2-2x-x+2=0
x(x-2) -1(x-2)
(x-2) (x-1)
x=2, x=1
The answer is A.