So I know I can say that $4a+b=11$ (by plugging in $2$), but I'm not sure where I can go from there. I've been looking for a second equation so I can do systems of equations, but I can't seem to find one. My thought was to isolate a, and say $a=\frac{11-b}{4}$ then say $\frac{11-b}{4}+b=11$, but that results in $b = 11$ and $a = 0$. I don't think that is correct. Please help.
2026-04-28 11:01:05.1777374065
$f(x)=ax^2+b$, if $f(x) $has a tangent line of $y=12x-13$ at $x=2$, find $a$ and $b$.
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First, what you have done is flawed. When you plug in the value of a in terms of b into the same equation, you have forgotten that the equation is actually $4a+b=11$ and not $a+b=11$. That's the reason you have got values for a and b when you actually shouldn't get any.
Now coming to the solution, $f'(x)$ gives you the slope of the curve at a certain point. So at the point x=2 the slope of the line is 4a(obtained through f'(x)). Now, from the equation of the tangent we get 4a=12. Therefore, a=3. b is suppose isn't a problem.