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!
Hello, BubblegumPops!
Determine $\displaystyle a,\,b,\text{ and }c$ so that the curve $\displaystyle y\:=\:ax^2+bx+c$ shall be tangent
to the line $\displaystyle 3x-y\:=\:5$ at (2,1) and shall also pass through (3,-1).
Since (2,1) is on the curve: .$\displaystyle a\!\cdot2^2 + b\!\cdot\!2 + c \:=\:1\quad\Rightarrow\quad{\color{blue}4a +2b + c \:=\:-1}$
Since (3,-1) is on the curve: .$\displaystyle a\!\cdot\!3^2 + b\!\cdot\!3 + c\:=\:-1\quad\Rightarrow\quad{\color{blue}9a + 3b + c \:=\:-1}$
The line $\displaystyle y \:=\:3x-5$ has slope 3.
The slope of the tangent to the curve: .$\displaystyle y' \:=\:2ax + b$
At $\displaystyle x=3\!:\;y'\;=\;2a\!\cdot\!3 + b \:=\:3\quad\Rightarrow\quad{\color{blue}6a + b \:=\:3}$
Solve the system of equations for $\displaystyle a,\,b,\text{ and }c.$