1. ## Derivatives and Tangents

Can someone describe how to solve this problem to me?

Determine a, b, and c so that the curve y=ax^2+bx+c shall be tangent to the line 3x-y=5 at (2,1) and shall also pass through (3,-1).

Thank you!

2. You need three conditions to uniquely determine a, b and c. We have:

y(x) = ax˛+bx+c, tangent line y = 3x-5 at (2,1) and point (3,-1).

So, the conditions are:

1) y(2) = 1
2) y'(2) = 3
3) y(3) = -1

Do you understand why? If so: solve the system.

3. Hello, BubblegumPops!

Determine $a,\,b,\text{ and }c$ so that the curve $y\:=\:ax^2+bx+c$ shall be tangent
to the line $3x-y\:=\:5$ at (2,1) and shall also pass through (3,-1).

Since (2,1) is on the curve: . $a\!\cdot2^2 + b\!\cdot\!2 + c \:=\:1\quad\Rightarrow\quad{\color{blue}4a +2b + c \:=\:-1}$

Since (3,-1) is on the curve: . $a\!\cdot\!3^2 + b\!\cdot\!3 + c\:=\:-1\quad\Rightarrow\quad{\color{blue}9a + 3b + c \:=\:-1}$

The line $y \:=\:3x-5$ has slope 3.
The slope of the tangent to the curve: . $y' \:=\:2ax + b$
At $x=3\!:\;y'\;=\;2a\!\cdot\!3 + b \:=\:3\quad\Rightarrow\quad{\color{blue}6a + b \:=\:3}$

Solve the system of equations for $a,\,b,\text{ and }c.$

4. hmm...i tried to solve the problem, but I don't think I got it right. I got a=15, b= -87, and c=125.

5. Originally Posted by BubblegumPops
hmm...i tried to solve the problem, but I don't think I got it right. I got a=15, b= -87, and c=125.
$4a+2b+c=-1$ ...[1]

$9a+3b+c=-1$ ...[2]

$6a+b=3$ ...[3]

[2] - [1]:

$5a+b=0$ ...[4]

[3] - [4]:

$a=3$

$\implies b=-15$

$\implies c = 17$

6. Thank you! I think i have it, but I don't understand how you got ...[1]. I thought that should equal 1, not -1.

7. Sorry, you're right. It should be $4a+2b+c=1$. Solving them similarly will yield $(a,b,c)=(5,-27,35)$

8. Thank you so much for your help!