- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,779

View attachment 1012

Let \(\displaystyle f(x)=\sqrt{x}\), Line \(\displaystyle L\) is the normal to the graph of

(a) show that the equation of

\(\displaystyle f'=\frac{1}{2\sqrt{x}}\) so \(\displaystyle f'(4)=\frac{1}{4}\) so normal would be \(\displaystyle -4\)

then

(b) Point \(\displaystyle A\) is the x-intercept of \(\displaystyle L\). find the x-coordinate of \(\displaystyle A\)

from \(\displaystyle Y=-4x+18\) set \(\displaystyle 0=-4x+18\) then \(\displaystyle 18=4x\) and \(\displaystyle x=\frac{9}{2}\) so \(\displaystyle A=(\frac{9}{2},0)\)

there is still (c) and (d) but want to make sure this is correct first.

Let \(\displaystyle f(x)=\sqrt{x}\), Line \(\displaystyle L\) is the normal to the graph of

*and point \(\displaystyle (4,2)\)***f**(a) show that the equation of

*is \(\displaystyle y=-4x+18\)***L**\(\displaystyle f'=\frac{1}{2\sqrt{x}}\) so \(\displaystyle f'(4)=\frac{1}{4}\) so normal would be \(\displaystyle -4\)

then

*is \(\displaystyle y-2=-4(x-4)\) or \(\displaystyle y= -4x+18\)***L**(b) Point \(\displaystyle A\) is the x-intercept of \(\displaystyle L\). find the x-coordinate of \(\displaystyle A\)

from \(\displaystyle Y=-4x+18\) set \(\displaystyle 0=-4x+18\) then \(\displaystyle 18=4x\) and \(\displaystyle x=\frac{9}{2}\) so \(\displaystyle A=(\frac{9}{2},0)\)

there is still (c) and (d) but want to make sure this is correct first.

Last edited: