# Figuring out values for a tangent equation on a curve

"What value of $b$ and $c$ for the curve $y = x^3 + bx^2 + cx$ give the tangent equation $y = 3x - 4$ in the point $(1, -1)$?"

This is what I do:

The derivative of the function is $3x^2 + 2bx + c$

In the point where $x = 1$, it becomes $3 + 2b + c$. Since the tangent equation is $3x - 4$, $3 + 2b + c = 3$. This means that $2b = -c$.

The answer in the book says that $c = -4$, and $b = 2$. However, shouldn't any value that conforms to $2b = -c$ work? I mean, they all give the slope of 3.

-

The question asks you to verify the equation of a tangent line to a given curve at a given point. That point must lie on the curve! You need $x=1$ and $y=-1$ to be a solution to the equation $y=x^3+bx^2+c$, i.e. you need $-1=1+b+c$ or equivalently: $b+c=-2$.

You can then do what you did and find the derivative and get the condition that $2b=-c$.

Solving the equation $b+c=-2$ and $2b=-c$ simultaneously gives $b=2$ and $c=-4$ as the book says.

-

You have to also make the original function pass through the point $(1,-1)$. That gives the equation

$$-1 = 1 + b + c,$$

which implies $b+c = -2$. Given $c = -2b$, as you have shown, this implies $-2 = b+c = b-2b = -b$, so indeed $b = 2$ and hence $c = -4$.

-
How does a function pass through at point? – Fly by Night Mar 14 '13 at 19:57
@FlybyNight The original equation $y=x^3+bx^2+cx$ must admit $(x,y) = (1,-1)$ as a valid solution. – gt6989b Mar 14 '13 at 20:01
You mean to say "the graph of the function". And even so, you can't make the graph of a function do anything; it goes where it goes. – Fly by Night Mar 14 '13 at 20:03

$c = -10, b = 5$ fulfills $2b = -c$, but $(1,-1)$ is not a solution to $y = x^3 + 5x^2 - 10x$

Remember, the line and the curve still have to meet at $(1, -1)$

-

you need to use the fact that $y=-1$ which means that $1 + b + c = -1$

-