Find the values of the constant 'a' that make the function \$f(x) = ax² + 3x\$ continuous at x = 2.

For a function to be continuous at a particular point, three conditions must be satisfied.

Direct substitution for f(2):

First, let’s find the value of the function at x = 2

\$ f(2) = a(2)^2 + 3(2) = 4a + 6 \$

Limit as x approaches 2:
Next, we need to find the limit of the function as x approaches 2

\$\lim_{x \to 2} {(ax^2 + 3x)} \$

Since this is a polynomial function, the limit is simply the function evaluated at x = 2

\$\lim_{x \to 2} {(ax^2 + 3x)} = a(2)^2 + 3(2) = 4a + 6\$

Continuity condition:
For the function to be continuous at x = 2, the above two expressions must be equal:

\$\lim_{x \to 2} f(x) = f(2) \$
4a + 6 = 4a + 6 (This is always true regardless of the value of a)

Therefore, there are no specific values of 'a' required for the function to be continuous at x = 2. It’s inherently continuous at that point for any value of 'a'.

This option accurately reflects the conclusion we reached. The continuity condition is independent of the specific value of 'a'

Any real number

This is incorrect. The value of "a" can be anything

a must be positive

This implies that 'a' needs to be a specific value (-3) for continuity. However, the limit and function hold true for any 'a', making this option incorrect

a = - 3

This option claims that only when 'a' is 0 will the function be continuous. However, like the previous options, it’s incorrect since 'a' doesn’t affect the continuity at x = 2

a = 0

Option

A

Can you summarize what you’ve understood in the above steps?