1. ## Tricky Differentiation Problem

Find the equation of the line having a negative slope, that passes through the point (2,-3), and is tangent to the curve y = x^2 + x

dy/dx = 2x^2 +1

Can someone point me in the right direction?

2. The equation of a line of slope $m$ passing through $(x_0,y_0)$ is

$y-y_0=m(x-x_0).$

Therefore, we seek a tangent line of $y=x^2+x$ of the form

$y-(x_0^2+x_0)=(2x_0+1)(x-x_0)$

that passes through $(2,-3)$. Here, we have used the fact that

$\frac{d}{dx}(x^2+x)=\frac{d}{dx}(x^2)+\frac{d}{dx} (x)=2x+1.$

Edit: Changed index to $0$ for clarity.

3. Originally Posted by millerst
Find the equation of the line having a negative slope, that passes through the point (2,-3), and is tangent to the curve y = x^2 + x

dy/dx = 2x^2 +1

Can someone point me in the right direction?
correction ... $\frac{dy}{dx} = 2x+1$

$y - (-3) = m(x - 2)$

$y + 3 = (2x+1)(x - 2)$

$(x^2+x)+3 = (2x+1)(x-2)$

$x^2+x+3 = 2x^2-3x-2$

$0 = x^2-4x-5$

solve for and choose the value of x that meets the given requirements.
I leave finding the tangent line equation to you.