Answer :
Sure! Let’s solve these polynomial functions step by step.
Let’s start with question 33: [tex]\( f(x) = x^2 - 36 \)[/tex].
### (a) Find all real zeros of the polynomial function
To find the zeros, we set the function equal to zero:
[tex]\[ x^2 - 36 = 0 \][/tex]
This is a difference of squares, which factors as:
[tex]\[ (x - 6)(x + 6) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ x - 6 = 0 \quad \text{or} \quad x + 6 = 0 \][/tex]
So the real zeros are:
[tex]\[ x = 6 \quad \text{and} \quad x = -6 \][/tex]
### (b) Determine whether the multiplicity of each zero is even or odd
Both zeros come from linear factors [tex]\((x - 6)\)[/tex] and [tex]\((x + 6)\)[/tex], which means each zero has a multiplicity of 1. Since 1 is odd, the multiplicities are:
- [tex]\( x = 6 \)[/tex]: odd multiplicity
- [tex]\( x = -6 \)[/tex]: odd multiplicity
### (c) Determine the maximum possible number of turning points of the graph of the function
The degree of the polynomial is 2 (since it is a quadratic function). The maximum number of turning points for a polynomial is one less than its degree:
[tex]\[ \text{Maximum turning points} = 2 - 1 = 1 \][/tex]
### (d) Verify using a graph (conceptually)
The graph of [tex]\( f(x) = x^2 - 36 \)[/tex] is a parabola that opens upwards with its vertex at the origin. It crosses the x-axis at [tex]\( x = 6 \)[/tex] and [tex]\( x = -6 \)[/tex], which matches our solutions for the zeros.
This procedure can be applied similarly to the other polynomial functions mentioned in your exercise list. If you have any other specific ones you'd like to walk through next, feel free to ask!
Let’s start with question 33: [tex]\( f(x) = x^2 - 36 \)[/tex].
### (a) Find all real zeros of the polynomial function
To find the zeros, we set the function equal to zero:
[tex]\[ x^2 - 36 = 0 \][/tex]
This is a difference of squares, which factors as:
[tex]\[ (x - 6)(x + 6) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ x - 6 = 0 \quad \text{or} \quad x + 6 = 0 \][/tex]
So the real zeros are:
[tex]\[ x = 6 \quad \text{and} \quad x = -6 \][/tex]
### (b) Determine whether the multiplicity of each zero is even or odd
Both zeros come from linear factors [tex]\((x - 6)\)[/tex] and [tex]\((x + 6)\)[/tex], which means each zero has a multiplicity of 1. Since 1 is odd, the multiplicities are:
- [tex]\( x = 6 \)[/tex]: odd multiplicity
- [tex]\( x = -6 \)[/tex]: odd multiplicity
### (c) Determine the maximum possible number of turning points of the graph of the function
The degree of the polynomial is 2 (since it is a quadratic function). The maximum number of turning points for a polynomial is one less than its degree:
[tex]\[ \text{Maximum turning points} = 2 - 1 = 1 \][/tex]
### (d) Verify using a graph (conceptually)
The graph of [tex]\( f(x) = x^2 - 36 \)[/tex] is a parabola that opens upwards with its vertex at the origin. It crosses the x-axis at [tex]\( x = 6 \)[/tex] and [tex]\( x = -6 \)[/tex], which matches our solutions for the zeros.
This procedure can be applied similarly to the other polynomial functions mentioned in your exercise list. If you have any other specific ones you'd like to walk through next, feel free to ask!
