# Thread: [SOLVED] Anyone know ?

1. ## [SOLVED] Anyone know ?

The curves y + ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).
Find a, b and c.

2. Originally Posted by jayson81
The curves y + ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).
Find a, b and c.
Firstly, (-1,0) is a common point thus.
Substitute that into the second function,
$0=2(-1)^2+c(-1)$
Thus,
$0=2-c$ solve, $c=2$
Thus, the second function has equation,
$y=2x^2+2x$.

Since we say that these two function have a common tangent means the equation of the tangent through point (-1,0) is the same. Remember from analytic geomtery that two lines having same slope at same point are the same. Same thing here the tangent lines share a point thus they need to have the same slope.

The slope at $x=-1$ for the first function is,
$y'=2ax$ at $x=-1$ is $-2a$.
The slope for the second is,
$y'=4x+2$ at $x=-1$ is $-2$
Thus, $-2=-2a$ thus, $a=1$
All we need is to find "b" in,
$y=x^2+b$ since (-1,0) is on the curve we have,
$0=(-1)^2+b$ thus, $b=-1$
Thus,
$(a,b,c)=(1,-1,2)$
Which is not true.
Thus, either you erred in the formulation of your question or there is no such numbers.

3. Hello, jayson81!

I got the same result as ThePerfectHacker, but I find that the solution works . . .

The curves $y = ax^2 +b$ and $y = 2x^2 +cx$ have a common tangent at the point $(-1,0)$.
Find $a$, $b$ and $c$.
Since $(-1,0)$ is on both curves, we have:
. . $a(-1)^2 + b \,=\,0\quad\Rightarrow\quad a + b \,=\,0$
. . $2(-1)^2 + c(-1)\,=\,0\quad\Rightarrow\quad c = 2$

Since the slopes are equal at $(-1,0)$, we have:
. . $2ax\,=\,4x + c\quad\Rightarrow\quad -2a\,=\,-4 + c$
Since $c = 2$, we get: $a = 1$ . . . and hence: $b = -1$

Therefore: . $(a,b,c)\,=\,(1,-1,2)$

The parabolas are: . $y\,=\,x^2 - 1$ and $y\,=\,2x^2 + 2x$
They share the common point $(-1,0)$ where they both have slope $-2.$