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?
The equation of a line of slope $\displaystyle m$ passing through $\displaystyle (x_0,y_0)$ is
$\displaystyle y-y_0=m(x-x_0).$
Therefore, we seek a tangent line of $\displaystyle y=x^2+x$ of the form
$\displaystyle y-(x_0^2+x_0)=(2x_0+1)(x-x_0)$
that passes through $\displaystyle (2,-3)$. Here, we have used the fact that
$\displaystyle \frac{d}{dx}(x^2+x)=\frac{d}{dx}(x^2)+\frac{d}{dx} (x)=2x+1.$
Edit: Changed index to $\displaystyle 0$ for clarity.
correction ... $\displaystyle \frac{dy}{dx} = 2x+1$
$\displaystyle y - (-3) = m(x - 2)$
$\displaystyle y + 3 = (2x+1)(x - 2)$
$\displaystyle (x^2+x)+3 = (2x+1)(x-2)$
$\displaystyle x^2+x+3 = 2x^2-3x-2$
$\displaystyle 0 = x^2-4x-5$
solve for and choose the value of x that meets the given requirements.
I leave finding the tangent line equation to you.