Question 1194922
<br>
{{{y=4sqrt(x+2)=4(x+2)^(1/2)}}}<br>
{{{dy/dt=(1/2)(4)(x+2)^(-1/2)=2/sqrt(x+2)}}}<br>
At (7,12), the slope of the tangent line is {{{2/sqrt(9)=2/3}}}, so the slope of the normal line is -3/2.<br>
Find the x-intercepts of the tangent line and normal line by setting y=0:<br>
{{{y-12=(2/3)(x-7)}}}
{{{-12=(2/3)(x-7)}}}
{{{-18=x-7}}}
{{{x=-11}}}<br>
{{{y-12=(-3/2)(x-7)}}}
{{{8=x-7}}}
{{{x=15}}}<br>
The two x-intercepts are -11 and 15, so the triangle has base length 15-(-11)=26 and height 12; its area is (1/2)(26)(12)=156<br>
ANSWER: 156<br>