pls explain?

Question

what is the area of the region bounded by the coordinate axes and the line tangent to the graph of y=(1/8)(x^2)+(1/2)x+1 at the point (0,1)

Answer 1

y = (1/8)(x^2) + (1/2)x + 1
y' = (1/4)x + 1/2
y'(0) = 1/2
y = x/2 + 1
y-intercept is 1
x-intercept is - 2
A = 1

Answer 2

I am not confident but will try.

y'=differentiation of (1/8)(x^2)+(1/2)x+1 w.r.t.x
y'=(1/8)2x + (1/2)*1+0
y'=(1/4)x +1/2

y' at (0,1) = slope of tangent m=1/2

Equation of line passing through (0,1) & slope 1/4
(y-1)=(1/2)(x-0)
y=x/2+1....this is the equation of the tangent at (0,1)

At y ordinate x=0 put this in tangent equation
y=0+1, y=1

At x ordinate y=0, put this in tangent equation
0=x/2+1
x=-2

Therefore, tangent will make a right angle triangle with coordinates having height as 1 unit (y=1 at y axis) & base as 2 unit (x=-2 at a axis).

The area of this triangle = (1/2)*Height*base
=(1/2)*(1)*(2)
=1 square unit

Answer 3

Jain made a slip ( but corrected it)

dy/dx=differentiation of (1/8)(x^2)+(1/2)x+1 w.r.t.x
dy/dx=(1/8)2x + (1/2)*1+0
dy/dx=(1/4)x +1/2
=> slope =(1/4)x +1/2
at x = 0
=> slope =(1/4)(0) +1/2
=> slope =1/2

EQT : y - y1 = m ( x - x1)
y-1= 1/2(x-0)
x - 2y + 2 = 0
cuts axis at
(0,1) (-2,0)

region = (0,0) (0,1) (-2,0)

Area of triangle = 1/2|x1 y2 – x2 y1|
=> 1/2 |(0)(0) - (-2)(1) |
=> 1 sq unit