MATEMATİK I

g(x) and p(x) are not in polynomial for because the powers of the elements must be natural numbers. Since -3 and 3/2 are not natural numbers, only f(x) and h(x) are polynomials.

h(x) is not a polynomial because, power of x in expression is -1, which is not a natural number. Besides f(x)=2 is a constant function, which means its degree is zero. Thus only g(x)=2x and k(x)=x/2 (which is equal to 0.5x) are polynomials of degree 1, namely they are linear functions.

-4/3, -3/4,-5/3 and -1 are not natural numbers, but 1 is. So the answer is E.

x²-5x+6=(x-2)(x-3)=0

Thus either x-2=0, which means x=2, or x-3=0, which means x=3. So the solution set is (2,3).

When you substitute x with zero, only the equation in C is true.

For a function's graph to pass through the origin, it must satisfy f(x)=0 for x=0. This condition is satisfied only for II and III

One parabola cross x axis only if it has real roots, which means in turn, the discirminant of parabola equation is greater than or equal to zero. This condition is satisfied only for the equation x²-5x+4

For a parabola to have a positive y intercept f(0) must be positive. This is true for only f(x)=x²-5x+4 for x=0 (since f(0)=4).

Firstly we have to find the roots of the quadratic expression f(x)=x²-5x+4=0

x²-5x+4=(x-1)(x-4)=0 is satisfied only if x=1 or x=4.

Thus we can divide the real numbers line to 3 regions:

1-)x<1

2-)1

3-) 4

In the first region (where x<1), f(x)=x²-5x+4>0. You can see this by substituting arbitrary numbers for x. (For instance think of x=0).

In the second region (where 12-5x+4<0. You can see this by substituting arbitrary numbers for x. (For instance think of x=2).

In the third region (where 42-5x+4>0. You can see this by substituting arbitrary numbers for x. (For instance think of x=5).

Thus only for the values of x in the second region, f(x) takes a negative value.

Firstly we have to find the roots of the quadratic expression f(x)=x²-6x-16=0

x²-6x-16=(x-8)(x+2)=0 is satisfied only if x=-2 or x=8.

Thus we can divide the real numbers line to 3 regions:

1-)x<-2

2-)-2

3-) 8

In the first region (where x<-2), f(x)=x²-6x-16>0. You can see this by substituting arbitrary numbers for x. (For instance think of x=-3).

In the second region (where -22-6x-16<0. You can see this by substituting arbitrary numbers for x. (For instance think of x=0).

In the third region (where 82-6x-16>0. You can see this by substituting arbitrary numbers for x. (For instance think of x=10).

Thus only for the values of x in the second region, f(x) takes a negative value.

 

This equation the sum of the roots is 5 and the product of the roots is 4. Therefore, the roots

should be 1 and 4.

if 7x-21=0 7x=21 and x=3.

if 4x-128<0 4x<128 and x<32

ÇK=(–∞, 32).

slope =a  for equation y = ax+b 

if 2y=-10x+4 y=-5x+2 and slope=-5

We find the common
solution. For this,  3x-14=-x-2 

                            3x+x=14-2

                           4x=12

                             x=3

Now we write x=3 in the equation y=-x-2=-3-2=-5  . So, (3,-5) is the intersection point of the lines