Consider the polynomials $$f(x) = ax^{2} + bx + c$$, where a > 0, b, c are real, and g(x) = -2x. If f(x) cuts the x-axis at (-2,0) and g(x) passes through (a, b), then the minimum value of f(x) + 9a + 1 is

$$g(x)=-2x$$, and it passes through (a,b) => $$b=-2a\rightarrow1$$

$$f(x) = ax^{2} + bx + c$$, and it cuts the x-axis at (-2,0) => $$0=a(4)+b(-2)+c$$. We will substitute the value of $$b=-2a$$ from eq. 1 => $$0=4a+4a+c$$ => $$c=-8a$$

=> $$f(x) = ax^{2} + (-2a)x + (-8a)$$

=> $$f(x) = ax^{2} -2ax -8a$$

minimum value of f(x) + 9a + 1 will be -

=> $$f(x)+9a+1 = ax^{2} -2ax -8a+9a+1$$

=> $$f(x)+9a+1 = ax^{2} -2ax + a+1$$

=> $$f(x)+9a+1 = a(x^{2} -2x + 1) + 1$$

=>$$f(x)+9a+1=a\left(x-1\right)^2+1$$

The minimum value of $$a(x-1)^2$$ is 0; thus, the minimum value of f(x) + 9a + 1 will be 1.