Let $f(x) = ax^2 + b$. Find $a, b$ so that linearization of $f(x)$ at $x = -2$ is $y = -4x - 7$

Question

I don't want you to solve the problem for me, but I'm am confused as to how to start the problem.

Linearization typically gives you the value of $a$, not $x$, with you then plugging that value into $L(x) = f(a) + f'(a)(x-a)$ and solving for $L(x)$.

In this case, I don't know if I should treat that $x$ value as '$a$' and plug it in for all the $a$'s in the $L(x)$ formula. And, the use of '$a$' in $f(x) = ax^2 + b$ adds some questions as well. Surely that's not the same '$a$' as the one in $L(x)$, because I assume that '$a$' is mostly arbitrary and stands for some number not $x$, that you will plug into the equation.

Answer 1

Hint: the lineariziation is given by:

$$L(x)=f'(-2)(x-(-2))+f(-2)$$ compare this with $L(x)=-4x-7$ to get a system of equations. Solving this system will determine $a$ and $b$.

Answer 2

"I don't want you to solve the problem for me, but I'm am confused as to how to start the problem. Linearization typically gives you the value of a, not x, with you then plugging that value into L(x)=f(a)+ f′(a)(x−a) and solving for L(x) In this case, I don't know if I should treat that x value as ' ' and plug it in for all the a's in the L(x)formula. And, the use of 'a' in $f(x)=ax^2+ b$ adds some questions as well. Surely that's not the same 'a' as the one in L(x), because I assume that 'a' is mostly arbitrary and stands for some number not x, that you will plug into the equation." First, 'x' is not any specific number, it is the variable in your function. Yes, 'a' is the x-value at which you are asked to find the tangent line. In this problem, you are told "at x= -2" so a= -2. Evaluate both f(-2) and f'(-2) and put them into your formula in place of f(a) and f'(a).