We have a function:
[tex]f(x)=x^3+12x^2+57x+14[/tex]
We have to find the point at which the slope of the tangent line is m = 9.
The slope of the tangent line of a function at a point is given by the value of the first derivative at that point, so we start by finding the first derivative:
[tex]\begin{gathered} f^{\prime}(x)=3x^2+12(2x)+57(1)+14(0) \\ f^{\prime}(x)=3x^2+24x+57 \end{gathered}[/tex]
We have now to find the value of x for which this derivative is equal to 9, so we can write:
[tex]\begin{gathered} f^{\prime}(x)=9 \\ 3x^2+24x+57=9 \\ 3(x^2+8x+19)=3(3) \\ x^2+8x+19=3 \\ x^2+8x+19-3=0 \\ x^2+8x+16=0 \end{gathered}[/tex]
We now have to find the roots of this equation to find the solution for x:
[tex]\begin{gathered} x=\frac{-8\pm\sqrt{8^2-4(1)(16)}}{2(1)} \\ x=\frac{-8\pm\sqrt{64-64}}{2} \\ x=\frac{-8\pm\sqrt{0}}{2} \\ x=-4 \end{gathered}[/tex]
This slope for the tangent line only happens for the point located at x = -4.
We can find the y-coordinate as y = f(-4):
[tex]\begin{gathered} f(-4)=(-4)^3+12(-4)^2+57(-4)+14 \\ f(-4)=-64+12(16)-228+14 \\ f(-4)=-64+192-228+14 \\ f(-4)=-86 \end{gathered}[/tex]
We can check this solution with a graph as:
Answer:
(-4, -86)
