Answer
$c=-\frac{1}{3}$ is the solution to the exercise.
Work Step by Step
$$(l): y=2x+3$$
The slope of line $(l)$ is $2$.
$$(S): y=f(x)=cx^2$$
- Derivative of $f(x)$: $$f'(x)=2cx$$
Line $(l)$ can only be tangent with curve $(S)$ if both of the followings happen:
1) Line $(l)$ intersects curve $(S)$ at at least one point $A(a, b)$.
2) At point A, $f'(a)$ equals the slope of line $(l)$.
*Consider 1):
If $A(a,b)$ is where line $(l)$ and curve $(S)$ intersects, then $A$ must lie in both the line and the curve. Therefore, $$b=ca^2\hspace{1cm}(1)$$ and $$b=2a+3\hspace{1cm}(2)$$
Consider 2): At point $A(a,b)$:$$f'(a)=2$$ $$2ca=2$$ $$ca=1$$ $$c=\frac{1}{a}\hspace{1cm}(3)$$
Substitute (3) into (1), we have $$b=\frac{1}{a}a^2=a\hspace{1cm}(4)$$
Substitue (4) into (2), we have $$a=2a+3$$ $$a=-3$$
Substitute $a=-3$ to (3), we have $$c=-\frac{1}{3}$$
$c=-\frac{1}{3}$ is the result.
