How to solve for f(1) in this function?

Question

Give that $f(x) = −4x^2 + x − 36$ for some real number $b$, part of which is graphed in the figure. What is the value of $f(1)$?

By plugging in $1$ for $x$ I come up with an expression: $-4 + b - 36$

What should be my next step for solving this problem?

Answer 1

Since $-3$ is a repeated root (from graph) of the quadratic $-4x^2+bx-36$, we have:

$$-4x^2+bx-36 = -4(x+3)^2$$

Comparison gives us $b=-24$.

Answer 2

From the graph it is clear that $\ f(x)$ has repeated root at $\ x = -3$ because at that point the quadratic touches x - axis .

So , $\ f'(x) = -8x + b$ , also has a root at $\ x = -3$ . Which implies $\ b = -24$ .

So , $\ f(x) = -4x^2-24x-36$ . And hence $\ f(1) = -64$.