Given
The derivative is given as
[tex]f^{\prime}(x)=4x+3[/tex]
and f(0)=-9.
Explanation
To find the function f(x),
[tex]dy=(4x+3)dx[/tex]
Take the integral,
[tex]\begin{gathered} \int dy=\int(4x+3)dx \\ y=\frac{4x^2}{2}+3x+C \\ y=2x^2+3x+C \\ f(x)=2x^2+3x+C \end{gathered}[/tex]
It is given that f(0) = - 9 .
[tex]\begin{gathered} -9=2(0^2)+3(0)+C \\ C=-9 \end{gathered}[/tex]
Then the function is determined as
[tex]f(x)=2x^2+3x-9[/tex]
Answer
Hence the function f(x) is
[tex]2x^2+3x-9[/tex]
