Differentiate the cubic, and let the derivative equal zero. You want the gradient to be 0 when x=12. So for a general cubic, f(x)=ax^3+bx^2+cx+d => f'(x)=3ax^2+2bx+c
=>0=3a(12)^2+2b(12)+c <---------(1)
You've got three other points so you can solve for a,b,c,d.
Exactly what grannysmith said. You let the derivative be equal to zero.
To elaborate further:
For the quadratic, you have:
y = ax^2 + bx + c
y' = 2ax + b
To head due east, let y' = 0
2ax + b = 0
x = -b/(2a)
Substitute this in to the original equation to find the y value:
y = a*(-b/(2a))^2 + b*(-b/(2a)) + c
y = b^2 / (4a) - b^2 / (2a) + c
y = -b^2 / (4a) + c
y = (4ac - b^2) / (4a)
Given the point (12, -4), we can solve for a, b and c:
When x = 12, y = -4 and y' = 0:
12 = -b/(2a)
b/a = -24
b = -24a
-4 = (4ac - b^2) / (4a)
4ac - b^2 = -16a
ac - (b/2)^2 = -4a
Substitute b = -24a:
ac - 144a^2 = -4a
ac = 140a^2
a*(140a - c) = 0
a = 0 or 140a = c
We are assuming that a is not equal to zero, since it's a quadratic, not a linear equation.
Therefore 140a = c
So we have b = -24a and c = 140a
Now we use our original equation and the point (12, -4):
y = ax^2 + bx + c
y = ax^2 - 24ax + 140a
-4 = 144a - 288a + 140a
-4 = -4a
a = 1
b = -24(1) = -24
c = 140(1) = 140
So the equation is:
y = x^2 - 24x + 140
For the cubic, you do the same thing essentially, just with an extra equation and an extra variable.
For the logarithm question, you know that logarithms have a horizontal asymptote. That means that the curve will never end up horizontal, it will only approach zero gradient, so it will never head due east.
We can prove this algebraically as well:
Let y = a*ln(bx + c) + d
Try to solve for y' = 0:
ab / (bx + c) = 0
0*(bx + c) = ab
0*bx = ab - c
0x = (ab - c) / b
No value of x will turn 0x into (ab - c) / b
Therefore, this equation is inconsistent.
Therefore, the logarithmic equation will never have zero gradient.
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