Find a and b in quadratic equation

Question

I have the problem to find $a$ and $b$ given

$f(x)=-x^2-2ax+b, a\neq0$

$f(1)=3$ , and the maximum value of $f(x)$ is $4$

and have they key with the answer $a=-2,b=0$, but which steps do I take to get to that answer?

Answer 1

$$f(1) = -1 - 2a + b = 3 \implies -2a + b = 4$$ and the maximum occurs when $f'(x)=0$, so $$-2x -2a = 0 \implies x = -a$$ We then know that $$ f(-a) = -(-a)^2 -2a(-a)+b=4 \implies a^2+b=4 $$ Equating our two equations, $$a^2+b = 4 = -2a+b \implies a=-2 \implies b=0$$

Answer 2

We complete the square to get $$f(x)=-(x+a)^2+(a^2+b)$$ If $f(1)=3$ then $3=-(1+a)^2+(a^2+b)$. The maximum value of $f(x)$ is for $x=-a$, when it is $a^2+b=4$. Now solve these two equations for $a,b$.

Answer 3

$f(x) = -(x + a)^2 + a^2 + b$. From this $f_{max} = 4 \to a^2 + b = 4$, and $f(1) = 3 \to -1 - 2a + b = 3 \to b = 2a + 4 \to a^2 + 2a + 4 = 4 \to a(a + 2) = 0 \to a = 0$ or $a = -2$. Since $a \neq 0$, $a = -2$, and so $b = 2a + 4 = 2\cdot (-2) + 4 = 0$