Question

Find an equation of a quartic function whose graph passes through the points (0, -2) and is tangent to the x-axis at (-1, 0) and (2, 0). (Leave answer in factored form)

Answer 1

The equation of the quartic function is $y = -\frac{1}{2}\cdot x^{4}+x^{3} +\frac{3}{2}\cdot x^{2} -2\cdot x - 2$, whose factored form is $y = (x-2)^{2}\cdot (x+1)^{2}$.

A quartic function is a fourth order polynomial, whose form is presented below:

$y = a\cdot x^{4}+b\cdot x^{3} + c\cdot x^{2}+d\cdot x + e$, $a, b, c, d, e\in \mathbb{R}$  (1)

Where:

• $x$ - Independent variable.
• $y$ - Dependent variable.

Given that such polynomial must be tangent to the x-axis in two place, slope must be zero and by differential calculus we have the following expression for the slope of the tangent line:

$4\cdot a \cdot x^{3} + 3\cdot b\cdot x^{2} + 2\cdot c \cdot x + d = 0$ (2)

Then, we can construct the following system of linear equations:

$e = -2$ (3)

$a -b +c -d = -e$ (4)

$-4\cdot a +3\cdot b -2\cdot c + d = 0$ (5)

$16\cdot a + 8\cdot b + 4\cdot c + 2\cdot d = -e$ (6)

$32\cdot a + 12\cdot b + 4\cdot c + d = 0$ (7)

The solution of the system is: $a = -\frac{1}{2} , b = 1, c = \frac{3}{2}, d = -2$.

The equation of the quartic function is $y = -\frac{1}{2}\cdot x^{4}+x^{3} +\frac{3}{2}\cdot x^{2} -2\cdot x - 2$, whose factored form is $y = (x-2)^{2}\cdot (x+1)^{2}$.