Geometrical Applications of Calculus Help! (1 Viewer)

[SGSII]

Hey guys!

Could you please help solve this question?

A cubic curve has a minimum turning point at (-2, -15), a maximum turning point at (1, 12) and its cuts the y axis at 5. What is the equation of the cubic?

I've never approached a question like this before :S
Please provide step by step solution.

Thanks

 

[Bob_Morris]

First off, we know the roots of the equation of the first derivative since we know the turning points.

Now we want an equation that will yield a minimum turning point at (-2,-15) and a maximum turning point at (1,12).
Such equation is y' = -a(x+2)(x-1) because if we drew this graph, then we can see that to the left of x=-2 is -ve and to the right is +ve which gives us a minimum turning point and for x=1 to the left is +ve and to the right is -ve which gives us a maximum turning point.

Now expanding y' = -a(x+2)(x-1) gives us:
y' = a(2-x-x²)

Since we want the equation, and we are dealing in the first derivative then we must integrate y' = 2-x-x²

Doing so,
y = a∫2-x-x² dx
y = a(2x - (x²/2) - (x³/3) + c)

Now to get find c, we must sub in a point in that we know exists on the curve.

We know that the curve cuts the y-axis at 5, so we sub (0,5)
5 = a[2(0) - (0²/2) - (0³/3) + c]
c = 5/a

Now to find a, sub in (1,12)
12 = a[2 - (1/2) - (1/3) + (5/a)]
12 = 2a - (a/2) - (a/3) + 5
12 = [12a - 3a - 2a + 30]/6
72 = 7a + 30
42 = 7a
a = 6

Therefore, the equation is
y = 6[2x - (x²/2) - (x³/3) + 5/6]

Expanding this will give us y= -2x³ - 3x² + 12x + 5

 

[SGSII]

First off, we know the roots of the equation of the first derivative since we know the turning points.

Now we want an equation that will yield a minimum turning point at (-2,-15) and a maximum turning point at (1,12).
Such equation is y' = -(x+2)(x-1) because if we drew this graph, then we can see that to the left of x=-2 is -ve and to the right is +ve which gives us a minimum turning point and for x=1 to the left is +ve and to the right is -ve which gives us a maximum turning point.

Now expanding y' = -(x+2)(x-1) gives us:
y' = 2-x-x²

Since we want the equation, and we are dealing in the first derivative then we must integrate y' = 2-x-x²

Doing so,
y = ∫2-x-x² dx
y = 2x - (x²/2) - (x³/3) + c

Now to get find c, we must sub in a point in that we know exists on the curve.

We know that the curve cuts the y-axis at 5, so we sub (0,5)
5 = 2(0) - (0²/2) - (0³/3) + c
c = 5

Therefore, the equation is
y = 2x - (x²/2) - (x³/3) + 5

Thanks!
But the answer given was y= -2x^3 - 3x^2 + 12x + 5
:S

 

[Bob_Morris]

sorry about that! i didn't take into consideration that it wasn't monic. fixed.

 

[SGSII]

LOL, its okays
Thanks for the answer!