Ok so im stuck on this and dont know how to approach it. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts.Here it is Write an equation for a cubic polynomial P(x)with leading coefficient −1 whose graph passes through the point (2, 8) and is tangent to the x axis at the origin.All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got the right answer from the choices but i did it by plugging in the point (2,8) and it came out to be -x^2(x-4) or -X^3+4x^2. Please help me as I have no clue on how to go about solving thanks.
The best way to solve a problem is to first list the conditions mathematically:
\(\displaystyle \text{1) } P(x) = -x^3 + a x^2+b x+c\)
\(\displaystyle \text{2) } P(2) = 8\)
\(\displaystyle \text{3) } P'(0) = 0\)
\(\displaystyle \text{4) } P(0) = 0\)
From \(\displaystyle \text{2)}\) we get \(\displaystyle -8 + 4a + 2b + c = 8\)
From \(\displaystyle \text{3)}\) we get \(\displaystyle -3(0)^2 + 2a(0) + b = 0\) and therefore \(\displaystyle b = 0\) and because of that \(\displaystyle 4a + c = 16\).
Now we can plug this into \(\displaystyle \text{1)}\) to get: \(\displaystyle P(x) = -x^3 + a x^2 - 4a + 16\) and from \(\displaystyle \text{4)}\) solve for a:
\(\displaystyle P(0) = -4a + 16 = 0 \to a = 4\)
Plug this back into \(\displaystyle \text{1)}\) to get: \(\displaystyle P(x) = -x^3 + 4 x^2\)
You can also go the easy way out by making this assumptions:
1) Any polynomial will be tangent at point (a,0) if and only if the multiplicity of a as a root of the polynomial is greater than 1.
Then we know the polynomial will be in this form:
\(\displaystyle P(x) = -x^2(x+c)\)
Now solve for \(\displaystyle c\) in \(\displaystyle P(2) = 8\)
\(\displaystyle 8 = -4(2+c) \to c = -4\)
And we get our same polynomial \(\displaystyle P(x) = -x^2(x-4) = -x^3 + 4 x^2\)
