given the following function find,
a. zeros and the multiplicity of each
b. number of turning points
c. end behavior
f(x)=(x-1)(x-5)
a. x=1 and x=5--is this correct?
b. number of turning points
I had f(x)-0 BUT my teacher said this--
Note that the equation has a degree of 2. f(x)=x^2-6x+5. How many turning points must an equation have that is of degree two? Is it two points--(1,0)and (5,0)
c. end behavior
I had for a large |x|(positiveor negative) f(x) increases as parabola upward to infinity BUT my teacher said not to change to an absolute value. Can someone help me with this one.
**please show work**
Your function is a simple parabola
All parabolas have one turning point, called the vertex
because of the + understood in front of
f(x) = +x^2 ....
the parabola opens upwards so it rises to infinity in both the 1st and 2nd quadrants.
If the function had been
f(x) = -x^2 .... , then it would have opened downwards. Your teacher was right to object to the |x| , that would always make the sign in front a +
Your zeros of x=1 and x=5 are correct, their multiplicity is 1
a. To find the zeros and their multiplicity, we need to set the function equal to zero:
f(x) = (x-1)(x-5) = 0
To solve this equation, we set each factor equal to zero:
x - 1 = 0 => x = 1
x - 5 = 0 => x = 5
Therefore, the zeros of the function are x = 1 and x = 5. Each zero has a multiplicity of 1 in this case because each factor is linear.
b. The number of turning points for a quadratic function (degree 2) is exactly one. So, your teacher is correct. The turning point occurs at the vertex of the parabola and can be found using the formula x = -b / (2a), where a and b are the coefficients of the quadratic function.
In this case, f(x) = x^2 - 6x + 5, so a = 1, and b = -6. Plugging these values into the formula, we get:
x = -(-6) / (2*1) = 6 / 2 = 3
So the turning point is at (3, f(3)).
c. To determine the end behavior of the function, we look at the leading term of the polynomial, which is the term with the highest exponent. In this case, the leading term is x^2.
If the leading term is positive, it means that as x approaches positive or negative infinity, f(x) also approaches positive infinity. Similarly, if the leading term is negative, f(x) approaches negative infinity.
In this case, the leading term coefficient is positive (1), so the end behavior of the function is as follows:
- As x approaches positive infinity, f(x) also approaches positive infinity.
- As x approaches negative infinity, f(x) also approaches positive infinity.
No absolute values are necessary in this case.
Therefore, the correct answers are:
a. Zeros and multiplicities: x = 1 (multiplicity 1), x = 5 (multiplicity 1)
b. Number of turning points: 1 turning point at (3, f(3))
c. End behavior: f(x) increases as x approaches positive or negative infinity.