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A cubic curve has a minimum turning point (-2,-15) and a maximum turning point of (1,12) and it cuts the y-axis at 5. What is the equation of the cubic?
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- Apr 12, 2005
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f(x) = a*x^3 + b*x^2 + c*x + d
"cuts the y-axis at 5"
f(0) = a*0^3 + b*0^2 + c*0 + d = d = 5
f(x) = a*x^3 + b*x^2 + c*x + 5
point (-2,-15) and point (1,12)
f(-2) = a*(-2)^3 + b*(-2)^2 + c*(-2) + 5 = -15
f(1) = a*(1)^3 + b*(1)^2 + c*(1) + 5 = 12
f'(x) = 3*a*x^2 + 2*b*x + c
Turning point (-2,-15) and Turning point (1,12)
f'(-2) = 3*a*(-2)^2 + 2*b*(-2) + c = 0
f'(1) = 3*a*(1)^2 + 2*b*(1) + c = 0
That looks like plenty of information.
"cuts the y-axis at 5"
f(0) = a*0^3 + b*0^2 + c*0 + d = d = 5
f(x) = a*x^3 + b*x^2 + c*x + 5
point (-2,-15) and point (1,12)
f(-2) = a*(-2)^3 + b*(-2)^2 + c*(-2) + 5 = -15
f(1) = a*(1)^3 + b*(1)^2 + c*(1) + 5 = 12
f'(x) = 3*a*x^2 + 2*b*x + c
Turning point (-2,-15) and Turning point (1,12)
f'(-2) = 3*a*(-2)^2 + 2*b*(-2) + c = 0
f'(1) = 3*a*(1)^2 + 2*b*(1) + c = 0
That looks like plenty of information.