m = 3x^3 + (1/2)x^-1/2 - 4/x^2
at the point x=1 this becomes:
m = 3(1)^3 + (1/2)(1)^-1/2 - 4/(1)^2
1 to any power is 1 so this simplifies to
m = 3(1) + (1/2)(1) - 4/1
m = 3 + 1/2 - 4
m = -1/2
now we just need a and b
we know that x=1 as it was given, so a=1
for b (y) at this point on the line, we just need to put x=1 into the original equation
b = x^3 + x^1/2 + (x^2+4x/x^2)
b= (1)^3 + (1)^1/2 + ((1)^2+4(1)/(1)^2)
b= (1 + 1 + (1+4/1)
b = 1 + 1 + 5
b = 7
so from the line equation: y-b = m (x-a)
where we now know m, a and b, we can subs in again and get:
y-7 = -1/2(x-1)
most teachers accept this answer, but you may want to simplify by solving for a
multiply out brackets
y-7 = -x/2 + 1/2
add 7 to both sides
y = 7.5 - x/2
PART B
First find the points A and B
A cuts the y axis, ie x=0
so y = 7.5 - 0/2
y = 7.5
the coordinates for this point are then A = (0,7.5)
and also B cuts the x axis, ie y=0
0 = 7.5 - x/2
x/2 = 7.5
x = 15
so coordinates are B = (15, 0)
and we know O = (0,0)
do a quick sketch of the triangle OAB and you should see its a simple right angle triangle with height 7.5 and base-length 15
the area for a triangle is 1/2(b*h), so (7.5*15)/2 = 56.25