Turning point of the curve

• Mar 27th 2010, 07:42 AM
mastermin346
Turning point of the curve
The curve $\displaystyle y=hx^2+\frac{k}{x}$ has a turning point at $\displaystyle P(\frac{1}{2},6)$.Find

(i)the value of $\displaystyle h$ and $\displaystyle k$

(ii)the equation of tangent to the curve at $\displaystyle x=1$
• Mar 27th 2010, 08:29 AM
skeeter
Quote:

Originally Posted by mastermin346
The curve $\displaystyle y=hx^2+\frac{k}{x}$ has a turning point at $\displaystyle P(\frac{1}{2},6)$.Find

(i)the value of $\displaystyle h$ and $\displaystyle k$

(ii)the equation of tangent to the curve at $\displaystyle x=1$

the point $\displaystyle P\left(\frac{1}{2} , 6 \right)$ is on the curve ...

$\displaystyle 6 = \frac{h}{4} + 2k$

since $\displaystyle P$ is a turning point ...

$\displaystyle y'\left(\frac{1}{2}\right) = 0$

use the above equation to get a second equation in terms of $\displaystyle h$ and $\displaystyle k$ ... then solve the system of equations.
• Mar 27th 2010, 08:42 AM
mastermin346