Finding the Area Inside Four Tangents for a Cubic Function | Step-by-Step Guide

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Homework Statement


Ok, I have a cubic function, y=(x-6)(x-1)(x-9) or y = x3-16x2+69x-54
I then have four tangents to make a quadrilateral, the tangents are as follows.
y=7.75x+14.75
y=-6.25x+56.25
y=7.75x + -76.231
y=-6.25x + 31.602
I need to find the area inside the tangents, except I have no idea where to start, Thankyou for any help in advance.
 
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jahaddow said:

Homework Statement


Ok, I have a cubic function, y=(x-6)(x-1)(x-9) or y = x3-16x2+69x-54
I then have four tangents to make a quadrilateral, the tangents are as follows.
y=7.75x+14.75
y=-6.25x+56.25
y=7.75x + -76.231
y=-6.25x + 31.602
I need to find the area inside the tangents, except I have no idea where to start, Thankyou for any help in advance.

Start by sketching the lines. What do you notice about these pairs of lines:

y=7.75x+14.75 & y=7.75x + -76.231

y=-6.25x+56.25 & y=-6.25x + 31.602

What sort of quadrilateral figure is formed?
 
A Parallelogram, But I don't see how I can get the area from the equations?
 
jahaddow said:
A Parallelogram, But I don't see how I can get the area from the equations?

You can find where the lines intersect one another and thus find the coordinates of the corners. You can then find the lengths of the lines. Split the area into two triangles and find the areas.
 
So, How do I do that?
 
jahaddow said:
So, How do I do that?

Solve each pair of equations to get the points of intersection. Find the lengths of the lines and then find the area of the shape.

Or you could split the area into two triangles and calculate it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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