Answer :
To find the equation of a cubic function [tex]\( f(x) \)[/tex] that passes through the points (3, 0) and (1, 4) and is tangent to the x-axis at the origin, let's follow these steps:
1. Form of the cubic equation:
Since the function is tangent to the x-axis at the origin, the origin (0, 0) is both a root and a point of inflection. Thus, the function must be in the form:
[tex]\[ f(x) = a x^3 + b x^2 + c x \][/tex]
Additionally, since it is tangent to the x-axis at [tex]\(x = 0\)[/tex], the coefficient [tex]\(d = 0\)[/tex].
2. Tangency condition at the origin:
Because it's tangent to the x-axis at the origin, [tex]\( x = 0 \)[/tex] is a root with a multiplicity of at least 2. The simplest way to ensure this is for [tex]\( f(x) \)[/tex] to contain [tex]\( x^2 \)[/tex] as a factor:
[tex]\[ f(x) = x^2 (a x + b) \][/tex]
We can rewrite this as:
[tex]\[ f(x) = a x^3 + b x^2 \][/tex]
3. Using given points to find coefficients:
We now use the points (3, 0) and (1, 4):
- Point (3, 0): [tex]\( f(3) = 0 \)[/tex]
[tex]\[ f(3) = a (3)^3 + b (3)^2 = 27a + 9b = 0 \Rightarrow 3a + b = 0 \][/tex]
- Point (1, 4): [tex]\( f(1) = 4 \)[/tex]
[tex]\[ f(1) = a (1)^3 + b (1)^2 = a + b = 4 \][/tex]
4. Solving the system of equations:
We have the system:
[tex]\[ \begin{cases} 3a + b = 0 \\ a + b = 4 \end{cases} \][/tex]
Solving for [tex]\( b \)[/tex] in the first equation:
[tex]\[ b = -3a \][/tex]
Substitute [tex]\( b \)[/tex] into the second equation:
[tex]\[ a - 3a = 4 \][/tex]
[tex]\[ -2a = 4 \][/tex]
[tex]\[ a = -2 \][/tex]
Now, find [tex]\( b \)[/tex]:
[tex]\[ b = -3(-2) = 6 \][/tex]
5. Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] back into the cubic function:
The cubic equation is:
[tex]\[ f(x) = -2x^3 + 6x^2 \][/tex]
Factored form:
[tex]\[ f(x) = x^2(-2x + 6) \][/tex]
[tex]\[ f(x) = -2x^3 + 6x^2 \][/tex]
[tex]\[ = -2x(x-3)x \][/tex]
So the factored form of the cubic function is:
[tex]\[ f(x) = -2x^2(x - 3) \][/tex]
This is the required cubic function whose graph passes through the points (3, 0) and (1, 4) and is tangent to the x-axis at the origin.
1. Form of the cubic equation:
Since the function is tangent to the x-axis at the origin, the origin (0, 0) is both a root and a point of inflection. Thus, the function must be in the form:
[tex]\[ f(x) = a x^3 + b x^2 + c x \][/tex]
Additionally, since it is tangent to the x-axis at [tex]\(x = 0\)[/tex], the coefficient [tex]\(d = 0\)[/tex].
2. Tangency condition at the origin:
Because it's tangent to the x-axis at the origin, [tex]\( x = 0 \)[/tex] is a root with a multiplicity of at least 2. The simplest way to ensure this is for [tex]\( f(x) \)[/tex] to contain [tex]\( x^2 \)[/tex] as a factor:
[tex]\[ f(x) = x^2 (a x + b) \][/tex]
We can rewrite this as:
[tex]\[ f(x) = a x^3 + b x^2 \][/tex]
3. Using given points to find coefficients:
We now use the points (3, 0) and (1, 4):
- Point (3, 0): [tex]\( f(3) = 0 \)[/tex]
[tex]\[ f(3) = a (3)^3 + b (3)^2 = 27a + 9b = 0 \Rightarrow 3a + b = 0 \][/tex]
- Point (1, 4): [tex]\( f(1) = 4 \)[/tex]
[tex]\[ f(1) = a (1)^3 + b (1)^2 = a + b = 4 \][/tex]
4. Solving the system of equations:
We have the system:
[tex]\[ \begin{cases} 3a + b = 0 \\ a + b = 4 \end{cases} \][/tex]
Solving for [tex]\( b \)[/tex] in the first equation:
[tex]\[ b = -3a \][/tex]
Substitute [tex]\( b \)[/tex] into the second equation:
[tex]\[ a - 3a = 4 \][/tex]
[tex]\[ -2a = 4 \][/tex]
[tex]\[ a = -2 \][/tex]
Now, find [tex]\( b \)[/tex]:
[tex]\[ b = -3(-2) = 6 \][/tex]
5. Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] back into the cubic function:
The cubic equation is:
[tex]\[ f(x) = -2x^3 + 6x^2 \][/tex]
Factored form:
[tex]\[ f(x) = x^2(-2x + 6) \][/tex]
[tex]\[ f(x) = -2x^3 + 6x^2 \][/tex]
[tex]\[ = -2x(x-3)x \][/tex]
So the factored form of the cubic function is:
[tex]\[ f(x) = -2x^2(x - 3) \][/tex]
This is the required cubic function whose graph passes through the points (3, 0) and (1, 4) and is tangent to the x-axis at the origin.
