http://img29.imageshack.us/img29/1954/capturend.png

Maybe we can walk through this? At first glance, the first thing I would do is find the derivative of $\displaystyle y=5x+45$, but that's not right.

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- Feb 4th 2010, 04:45 PMBeSweeetFinding k of a Tangent Line
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Maybe we can walk through this? At first glance, the first thing I would do is find the derivative of $\displaystyle y=5x+45$, but that's not right. - Feb 4th 2010, 04:50 PMpickslides
- Feb 4th 2010, 04:57 PMBeSweeet
- Feb 4th 2010, 05:03 PMpickslides
- Feb 4th 2010, 05:08 PMBeSweeet
- Feb 5th 2010, 03:46 AMHallsofIvy
If you haven't learned "the whole dy/dx thing", why did you refer to finding the derivative in your first post? The formula you give is for the slope of a straight line. If f(x) is not linear, it is the slope of a "secant" line, a line that crosses the graph at x and x+y, not the tangent line. Since you don't yet know the derivative (I suspect this problem is preliminary to introducing the derivative), here's a method that predates the calculus:

If y= 5x+ 45 and $\displaystyle y= k\sqrt{x}$ meet at all, we must have $\displaystyle y= 5x+ 45= k\sqrt{x}$ or $\displaystyle 5x- k\sqrt{x}+ 45= 0$. Let u= $\displaystyle \sqrt{x}$ so that $\displaystyle x= u^2$. The equation becomes $\displaystyle 5u^2- ku+ 45= 0$. If x= a at the point of intersection then x- a= $\displaystyle u- \sqrt{a}$ must be a factor of that polynomia. If fact, to be**tangent**that must be a**double**factor- we must have $\displaystyle 5u^2- ku+ 45= 5(u- \sqrt{a})^2$ Multiplying the right side, we get $\displaystyle 5u^2- ku+ 45= 5u^2- 10\sqrt{a} u+ 5a$ for all u. Comparing coefficients, $\displaystyle -k= -10\sqrt{a}$ and $\displaystyle 5a= 45$. Solve the second equation for a and use that to solve the first equation for k.