Question

Hello, please help ASAP. Thank you!

Answer 1

Answer:

23) No

24) No

25) Yes

Step-by-step explanation:

Question 23)

We want to determine if a zero exists between 1 and 2 for the function:

$f(x)=x^2-4x-5$

Find the zeros of the function. We can factor:

$\displaystyle 0 = (x-5)(x+1)$

Zero Product Property:

$x-5=0\text{ or } x+1=0$

Solve for each case. Hence:

$\displaystyle x = 5\text{ or } x=-1$

Therefore, our zeros are at x = 5 and x = -1.

In conclusion, a zero does not exist between 1 and 2.

Question 24)

We have the function:

$f(x)=2x^2-7x+3$

And we want to determine if a zero exists between 1 and 2.

Factor. We want to find two numbers that multiply to (2)(3) = 6 and that add to -7.

-6 and -1 suffice. Hence:

$\displaystyle \begin{aligned} 0 & = 2x^2-7x + 3 \\ & = 2x^2 -6x -x + 3 \\ &= 2x(x-3) - (x-3) \\ &= (2x-1)(x-3) \end{aligned}$

By the Zero Product Property:

$2x-1=0\text{ or } x-3=0$

Solve for each case:

$\displaystyle x=\frac{1}{2} \text{ or } x=3$

Therefore, our zeros are at x = 1/2 and x = 3.

In conclusion, a zero does not exist between 1 and 2.

Question 25)

We have the function:

$f(x)=3x^2-2x-5$

And we want to determine if a zero exists between -2 and 3.

Factor. Again, we want to find two numbers that multiply to 3(-5) = -15 and that add to -2.

-5 and 3 works perfectly. Hence:

$\displaystyle \begin{aligned} 0&= 3x^2 -2x -5 \\ &= 3x^2 +3x - 5x -5 \\ &= 3x(x+1)-5(x+1) \\ &= (3x-5)(x+1)\end{aligned}$

By the Zero Product Property:

$\displaystyle 3x-5=0\text{ or } x+1=0$

Solve for each case:

$\displaystyle x = \frac{5}{3}\text{ or } x=-1$

In conclusion, there indeed exists a zero between -2 and 3.