How do I find ordered pair, given slope of the tangent line? The function is $f(x) = x^3 + 9x^2 + 36x + 10$ and the slope given is $9$.
I found the derivative and set it equal to $9$, but I ended up with $x = (-9,-33)$ and the answer is $(-3,-44)$.
I've asked two Math majors and neither knows how to find it.
Where did I go wrong and how can I answer the next one correctly?
Work:
\begin{align*}
& f(x) = x^3 + 9x^2 + 36x + 10 \Rightarrow f^{\prime}(x) = 3x^2 + 18x + 36 \Rightarrow 3x^2 + 18x + 36 = 9 \Rightarrow\\\\
& 3x^2 + 18x = -27 \Rightarrow 3x ( x + 6 ) = -27 \Rightarrow 3x = -27  x + 6 = -27 \Rightarrow
x = -3  x = -33
\end{align*}
 A: So we know the derivative value (i.e. tangent slope gradient), but not the point on $f(x)$. 
First derive and set derivative to 9. 
$ f'(x) = 3x^{2}+18x+36=9 $ 
$ \Rightarrow x^{2}+6x+12-3=0 $           (divide by 3 then subtract 3) 
$ \Rightarrow (x+3)^{2}=0 $ 
$ \therefore x=-3$           ($x$-ordinate of the point on f(x)) 
To find the $y$-oordinate substitute $x$ back into $f(x)$: 
$f(-3)=-44$ 
Hence at the point $(-3,-44)$ the function $f(x)$ has a tangent slope gradient of $9$.
A: You got $3x^2+18x=-27.$ 
This is equivalent to $3x^2+18x+27=0$ or $x^2+6x+9=0$  or $(x+3)^2=0$.
Can you take it from here?
A: From your second line at start you get by simplification for given slope $9$
$$ (x+3)^2 =0$$
Plug $x=-3$ into $y=f(x)$ expression and you directly get $ y=-44$ so that this point ( called inflection point ) is described as
$$ y= f(x)= x^3+9x^2+36x+10 $$
$$\frac{dy}{dx}= 3x^2+18x +36 $$
$$ \frac{d^2y}{dx^2}=6x+18 $$
$$ \frac{d^3y}{dx^3}=6 $$
$$ x=-3,y=-44,\frac{dy}{dx}=9, \frac{d^2y}{dx^2}=0  $$
all of which is seen in the WA plot
