Determine the smallest integer value of a for which f(x) has imaginary zeroes. Show how you found
this answer.

The question is incomplete, the complete question is

Determine the smallest integer value of a for which f(x) = ax² - 2x + 5 has imaginary zeroes. Show how you found  this answer

The smallest integer value of a is 1

Step-by-step explanation:

To find the zeroes of a function equate it by 0, then find the values of x which are the zeroes of the function

To find the types of the roots (zeroes) of a function f(x) = ax² + bx + c use the discriminant of the function b² - 4ac

∵ f(x) = ax² - 2x = 5

- To find its zeroes equate f(x) by 0

∴ ax² - 2x + 5 = 0

∵ f(x) has imaginary zeroes

- That means the discriminant is less then zero

∴ b² - 4ac < 0

∵ a = a, b = -2 and c = 5

- Substitute them in the inequality above

∴ (-2)² - 4(a)(5) < 0

∴ 4 - 20 a < 0

- Add 20 a to both sides

∴ 4 < 20 a

- Divide both sides by 20

∴ 0.2 < a

- That means a is greater than 0.2

∴ a > 0.2

∵ You must to find the smallest integer value of a

- The first integer greater than 0.2 is 1

∴ a = 1

The smallest integer value of a is 1

susie has 7 apples

(2,6)

step-by-step explanation:

cause i just did this and got the answer

you're welcome