How do I find ordered pair, given slope of the tangent line?

Question

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*}

Answer 1

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?

Answer 2

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

Answer 3

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 $9$.