Originally Posted by

**Jhevon** we know that the lines must be of the form y = mx + b, right? since they pass through the point (5,10), we have by the point-slope form that te lines are of the form:

$\displaystyle y - 10 = m(x - 5)$

$\displaystyle \Rightarrow y = mx + (10 - 5m)$

Now, since the lines are tangent to the quadratic, their slopes at some point must be the same as the quadratic's and hence can be given by the same formula.

as you rightly said, the slope of the quadratic is given by the formula $\displaystyle f'(x) = 10 - 2x$, so we can set this for m in the equation of the line above. thus we have:

$\displaystyle y = (10 - 2x)x + (10 - 50 + 10x)$

$\displaystyle \Rightarrow y =- 2x^2 + 20x - 40$

since the lines touch the curve at one point, we equate the formulas to find what x's the lines touch the curve

thus we have: $\displaystyle -2x^2 + 20x - 40 = -16 + 10x - x^2$

I think you can take it from here