3. the graph of a function F crosses the x-axis at -1 and 3 and touches the x-axis at 5. which equations could define this function select all that apply
a- f(x)= (x+1)(x-3)(x-5)
b- f(x)= (x+1)(x-3)(x-5)^2
c- f(x)= (x+1)^4(x-3)^2(x-5)^5
d- f(x)= (x+1)^7(x-3)^3(x-5)^4
4. which polynomial functions have exactly one positive real zero select all that apply
a- f(x)= 7x^4-5x^3-4x+1
b- f(x)= 3x^4-10x^3-x-7
c- f(x)= -x^5+4x^3-2x^2-3x+7
d- f(x)= 2x^5+8x^2+4x+1
3. B and D -- f(x)=(x+1)(x-3)(x-5)^5 and f(x)=(x+1)^7(x-3)^3(x-5)^4
4. B and C -- f(x)=3x^4-10x^3-x-7 and f(x)=-x^5+4x^3-2x^2-3x+7
:)
a- f(x)= (x+1)(x-3)(x-5)
b- f(x)= (x+1)(x-3)(x-5)^2
c- f(x)= (x+1)^4(x-3)^2(x-5)^5
No equation could define this function because it is constantly escaping from the x-axis. It's like trying to catch a slippery fish! The graph just skims the surface at 5, but doesn't make a lasting touch. So, none of the equations provided would be a suitable fit. It's like trying to eat a banana without peeling it first - it just won't work!
b- f(x)= 3x^4-10x^3-x-7
c- f(x)= -x^5+4x^3-2x^2-3x+7
These two polynomial functions have exactly one positive real zero. It's like finding a needle in a haystack, except in this case, you only have one needle to find! So, options b and c would be the correct answers here. Well done for narrowing it down!
For question 3, the graph of a function F crosses the x-axis at -1 and 3 and touches the x-axis at 5.
To determine which equations could define this function, we can look at the x-intercepts and the behavior of the graph near those points.
The function crosses the x-axis at -1 and 3, meaning the factors (x+1) and (x-3) must be present in the equation.
The function touches the x-axis at 5, which means the factor (x-5) must be squared in the equation.
Using this information, the equations that could define this function are:
a- f(x)= (x+1)(x-3)(x-5)
b- f(x)= (x+1)(x-3)(x-5)^2
For question 4, which polynomial functions have exactly one positive real zero, there should be only one real zero that is positive. This means the function should only cross or touch the x-axis at one point on the positive side.
Looking at the given polynomial functions, the functions that have exactly one positive real zero are:
c- f(x)= -x^5+4x^3-2x^2-3x+7
d- f(x)= 2x^5+8x^2+4x+1
The other functions have either more than one positive real zero or no positive real zeros at all.
To determine which equations could define the given functions, we can examine the provided information about the graph.
3. The graph of function F crosses the x-axis at -1 and 3 and touches the x-axis at 5.
We know that when a function crosses the x-axis, its corresponding factor is equal to zero. Therefore, the factors of the function F would be (x+1), (x-3), and (x-5).
Since the graph touches the x-axis at 5, but doesn't cross it, the factor (x-5) is raised to the power of at least 2.
Now, let's analyze the given options:
a- f(x) = (x+1)(x-3)(x-5)
This option includes the factors (x+1), (x-3), and (x-5), but it doesn't account for the touching point at 5. Therefore, this option is not applicable.
b- f(x) = (x+1)(x-3)(x-5)^2
This option includes the factors (x+1), (x-3), and (x-5)^2, which accounts for the touching point at 5. Therefore, this option could define the function F.
c- f(x) = (x+1)^4(x-3)^2(x-5)^5
This option includes higher powers of all the factors. It goes beyond the information given, so it is not necessary to include those higher powers. Therefore, this option is not applicable.
d- f(x) = (x+1)^7(x-3)^3(x-5)^4
Similar to option c, this option includes higher powers that are not necessary to define the function F. Therefore, this option is not applicable.
Hence, the equations that could define the function F are:
- f(x) = (x+1)(x-3)(x-5)^2 (option b)
Moving on to question 4, which asks for polynomial functions that have exactly one positive real zero:
To have exactly one positive real zero, a polynomial function must have only one positive root and possibly other complex or negative roots. This implies that in the factored form of the polynomial, there should only be one linear factor of the form (x-c), where c is positive, and all other factors would be of higher powers.
Let's analyze the given options:
a- f(x) = 7x^4 - 5x^3 - 4x + 1
This polynomial function does not have exactly one positive real zero, as it has multiple terms and powers. Therefore, this option is not applicable.
b- f(x) = 3x^4 - 10x^3 - x - 7
Similar to option a, this polynomial function does not have exactly one positive real zero, as it has multiple terms and powers. Therefore, this option is not applicable.
c- f(x) = -x^5 + 4x^3 - 2x^2 - 3x + 7
Although this polynomial function has a negative leading coefficient, it still has multiple terms and powers. Therefore, it does not have exactly one positive real zero. This option is not applicable.
d- f(x) = 2x^5 + 8x^2 + 4x + 1
This polynomial function has only one linear factor (x-1), making it the only candidate that satisfies the condition of having exactly one positive real zero.
Hence, the polynomial function that has exactly one positive real zero is:
- f(x) = 2x^5 + 8x^2 + 4x + 1 (option d)
#3.B because of the double root
#4. Because of the one real root, the degree must be odd. So, A and B are out.
By Descartes' Rule of Signs,
A has at most 2 positive and 0 negative roots
Now apply the Rule to C and D