I need some help, this is from a non-calculator wkst.
1) Find the equations of both lines through (2, -3) that are tangent to the parabola $\displaystyle y=x^2+x$
let $\displaystyle (x,y)$ be a point on the tangent line
then the slope between $\displaystyle (x,y)$ and $\displaystyle (2,-3)$ is given by $\displaystyle m = \frac {y_2 - y_1}{x_2 - x_1} = \frac {y + 3}{x - 2} = \frac {x^2 + x + 3}{x - 2}$
but this formula for the slope must be equal to the formula for the derivative. since the derivative also gives the slope of the tangent line. so that
$\displaystyle y' = 2x + 1 = \frac {x^2 + x + 3}{x - 2}$
solving $\displaystyle 2x + 1 = \frac {x^2 + x + 3}{x - 2}$, we find the solutions to be $\displaystyle x = 5$ or $\displaystyle x = -1$
so, the slopes of our two tangent lines are given by $\displaystyle y'(5) = 11$ and $\displaystyle y'(-1) = -1$
thus, our two lines are:
(1) the line passing through (2,-3) with slope 11
and
(2) the line passing through (2,-3) with slope -1
i leave it to you to find these lines
please, no bumping. it is against the rulesUrgent, please help.
also, do you go to CCNY?