Given the following polynomial find,
a. zeros and the multiplicity of each
b. number of turning points
c. end behavior
f(x)=(x-1)(x-5)
a. the multiplicity of (x-1)(x-5)
x=1 and x=5----is this correct
b. number of turning points
I had f(x)-0 at turning points BUT my teacher said note that the equation is a degree of two. f(x)=x^2-6x+5; how many turning points must an equation have that is of degree two?
c. end behavior
I had a large |x|(positive or negative) f(x)increases as a parabola upward to ionfinity. **My teacher said not ot change to an absolute value; so now I am really confused. Can someone help me please?
**please show work**
a. To find the zeros and multiplicities of the polynomial f(x) = (x-1)(x-5), we need to set the polynomial equal to zero and solve for x.
Setting f(x) = 0, we have (x-1)(x-5) = 0.
Using the zero product property, we can set each factor equal to zero:
x-1 = 0 --> x = 1
x-5 = 0 --> x = 5
Therefore, the zeros of the polynomial are x = 1 and x = 5. Each zero corresponds to a factor, and the multiplicity of each zero is 1. So, the multiplicity of (x-1)(x-5) is 1 for each factor.
b. The number of turning points in a polynomial is determined by the degree of the polynomial. In this case, the polynomial f(x) = (x-1)(x-5) is a degree 2 polynomial, also known as a quadratic.
Quadratic polynomials have a maximum of one turning point. Therefore, f(x) = (x-1)(x-5) can have at most 1 turning point. In this case, we can find the turning point by finding the vertex of the parabola, which occurs at the x-coordinate of the vertex of the quadratic equation.
The general form of a quadratic equation is f(x) = ax^2 + bx + c. In this case, a = 1, b = -6, and c = 5.
The x-coordinate of the vertex is given by x = -b/2a. Plugging in the values, we have x = -(-6)/(2*1) = 6/2 = 3.
Therefore, the quadratic polynomial has one turning point at x = 3.
c. To determine the end behavior of a polynomial, we need to look at the leading term of the polynomial.
In this case, the leading term is x^2, which has a positive coefficient of 1.
For large positive values of x, the value of x^2 will be positive, and as a result, the entire polynomial f(x) = (x-1)(x-5) will also be positive. This means that as x approaches positive infinity, f(x) increases without bound.
Similarly, for large negative values of x, the value of x^2 will still be positive, and consequently, f(x) = (x-1)(x-5) will also be positive. Thus, as x approaches negative infinity, f(x) increases without bound.
So, the end behavior of f(x) = (x-1)(x-5) is that as x approaches positive or negative infinity, f(x) increases without bound in a parabolic fashion.