Question

Please help me i'll mark as brainliest​

Answer 1

Answer:   a = 0

Step-by-step explanation:

ok we got $f(x) = x^2 + ax + 3$. Our task is to choose a value for $a$ so that "the tangent line to $f$ at $x = 2$ passes through $(1, 3)$."

Well $f(2) = 4 + 2a + 3 = 2a + 7$, and $f'(2) = 2(2) + a = 4 + a$. So our tangent line looks like this:

$y - 3 = (4 + a)(x - 1)$

$y = 4x + ax - a - 1$

We know this must contain the point $(2, 2a + 7)$, so

$2a + 7 = 4(2) + a(2) - a - 1$

$a = 8 - 7 - 1\\\\\boxed{a = 0}$

Answer 2

Just had this on my test! A=0 is 100% correct!