Tangent line parallel to another line At what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find its equation.
I don't know what the slope of the tangent line will be. Is it the negative reciprocal? 
 A: To be parallel, two lines must have the same slope.
The slope of the tangent line at a point of the parabola is given by the derivative of $y= x^2-3x-5$.
This means that the question is asking at what point the derivative of the parabola will equal the slope of $3x-y=2$.
So, to solve the problem, identify the slope of the line and set it equal to the derivative of the equation of the parabola to find the $x$ value of the point you want. Then use the equation of the parabola to find the $y$ value, and you're done.
A: $y=x^2-3x-5$ ,$dy/dx= 2x-3$ slope of the tangent line parallel to $3x-y=2$ ,whose slope is 3 which means $ 3=2x-3$ i.e. $x=3 $,$y=-5$ and the equation of tangent will be $y+5=3(x-3)$
A: Edit: since the tangent is parallel to the given line: $3x-y=2$  hence the slope of tangent line to the parabola is $\frac{-3}{-1}=3$ 
Let the equation of the tangent be $y=3x+c$ 
Now, solving the equation of the tangent line: $y=3x+c$ & the parabola: $y=x^2-3x-5$ by substituting $y=3x+c$ as follows $$3x+c=x^2-3x-5$$ $$\implies x^2-6x-(c+5)=0\tag 1$$ For tangency we have the following condition $$ \text{determinant},\ B^2-4AC=0$$ $$\implies (-6)^2-4(1)(-(c+5))$$ $$\implies c=\frac{-56}{4}=-14$$ Hence, setting the value of $c=-14$ we get $$x^2-6x-(-14+5)=0$$ $$\implies x^2-6x+9=0$$ $$\implies (x-3)^2=0\implies x=3$$ Now, setting the value of $x=3$ in the equation of parabola as follows $$y=(3)^2-3(3)-5=-5$$ Hence, the point of tangency is $\color{blue}{(3, -5)}$ 
A: If you solve simultaneously the curve and the line $y=3x+c$ to get a quadratic equation in $x$ then this quadratic must have double roots at the point of tangency. This will give the value of $c$ and the required $x$ value is given by $x=-\frac{b}{2a}$
A: Slope of  tangent to parabola $y=x^2-3x-5$ is parallel to slope of line $3x-y=2$ when the angle between them is $0$.
Given parabola
$$y=x^2-3x-5$$
Differentiating w.r.t. "x"
$$\frac{dy}{dx}=2x-3$$
Slope of tangent to parabola
$$m_1=\frac{dy}{dx}=2x-3$$
Given line
$$3x-y=2$$
Just to write in the form of $y=mx+c$
$$y=3x-2$$
So the slope of line is
$$m_2=3$$
Now to find the angle between two slope we have
$$tan\theta = \frac{m_2-m_1}{1+m_2m_1}$$
$$tan\theta = \frac{3-2x+3}{1+3(2x-3)}$$
Here we need to find the value of "$x$"
