T towlie536 New member Joined Jan 23, 2008 Messages 1 Jan 23, 2008 #1 A cubic polynomial function f has leading coefficient 2 and constant term -5. If f(1)=0 and f(2)=3, what is f(-5)? Explain how you found your answer. I dont get how to set this up or solve it. Can anyone help?
A cubic polynomial function f has leading coefficient 2 and constant term -5. If f(1)=0 and f(2)=3, what is f(-5)? Explain how you found your answer. I dont get how to set this up or solve it. Can anyone help?
D Deleted member 4993 Guest Jan 24, 2008 #2 Re: A cubic polynomial function.. Help! towlie536 said: A cubic polynomial function f has leading coefficient 2 and constant term -5. If f(1)=0 and f(2)=3, what is f(-5)? Explain how you found your answer. I dont get how to set this up or solve it. Can anyone help? Click to expand... A general cubic polynomial will look like: \(\displaystyle a\cdot x^3\, + \, b\cdot x^2\, + \, c\cdot x\, + \,d\) where a, b, c & d are unknown constants - that you need to find. Your problem tells you a = 2 and d = -5 Now you have to find two more constants (b & c). For that you are given two conditions (f(1) = 0 and f(2) =3). These will give you two equations and you solve for two unknowns. Please show us your work and exactly where you are stuck - so that we can help you better.
Re: A cubic polynomial function.. Help! towlie536 said: A cubic polynomial function f has leading coefficient 2 and constant term -5. If f(1)=0 and f(2)=3, what is f(-5)? Explain how you found your answer. I dont get how to set this up or solve it. Can anyone help? Click to expand... A general cubic polynomial will look like: \(\displaystyle a\cdot x^3\, + \, b\cdot x^2\, + \, c\cdot x\, + \,d\) where a, b, c & d are unknown constants - that you need to find. Your problem tells you a = 2 and d = -5 Now you have to find two more constants (b & c). For that you are given two conditions (f(1) = 0 and f(2) =3). These will give you two equations and you solve for two unknowns. Please show us your work and exactly where you are stuck - so that we can help you better.
