Answer: a = 0
Step-by-step explanation:
ok we got [tex]f(x) = x^2 + ax + 3[/tex]. Our task is to choose a value for [tex]a[/tex] so that "the tangent line to [tex]f[/tex] at [tex]x = 2[/tex] passes through [tex](1, 3)[/tex]."
Well [tex]f(2) = 4 + 2a + 3 = 2a + 7[/tex], and [tex]f'(2) = 2(2) + a = 4 + a[/tex]. So our tangent line looks like this:
[tex]y - 3 = (4 + a)(x - 1)[/tex]
[tex]y = 4x + ax - a - 1[/tex]
We know this must contain the point [tex](2, 2a + 7)[/tex], so
[tex]2a + 7 = 4(2) + a(2) - a - 1[/tex]
[tex]a = 8 - 7 - 1\\\\\boxed{a = 0}[/tex]
