Question

Find a cubic function \(y = a{x^3} + b{x^2} + cx + d\) whose graph has horizontal tangents at the points\(\left( { - 2,6} \right)\) and \(\left( {2,0} \right)\).

Short answer

The required cubic function is \(y = \frac{3}{{16}}{x^3} - \frac{9}{4}x + 3\).

Step-by-step solution

Cubic Polynomials

The cubic polynomial is a polynomial that has the value of its highest degree equal to 3, with the condition that the polynomial must consist of only one variable and real numbers as constants.

Evaluating equations using given data.

Thegiven polynomial is \(y = a{x^3} + b{x^2} + cx + d\)

Differentiating with respect to\(x\)as:

\(y'\left( x \right) = 3a{x^2} + 2bx + c\)

Now, for the points \(\left( { - 2,6} \right)\,\& \,\left( {2,0} \right)\), we have:

\(\begin{align}y\left| {_{x = - 2}} \right. &= 6\\ - 8a + 4b - 2c + d &= 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( {{\rm{eq}}{\rm{.1}}} \right)\\y\left| {_{x = 2}} \right. &= 0\\8a + 4b + 2c + d &= 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( {{\rm{eq}}{\rm{.2}}} \right)\end{align}\)

For horizontal tangents at points \(\left( { - 2,6} \right)\,\& \,\left( {2,0} \right)\), we have:

\(\begin{align}y'\left( x \right)\left| {_{x = - 2}} \right. &= 0\\12a - 4b + c &= 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( {{\rm{eq}}{\rm{.3}}} \right)\\y'\left( x \right)\left| {_{x = 2}} \right. &= 0\\12a + 4b + c &= 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,................\left( {{\rm{eq}}{\rm{.4}}} \right)\end{align}\)

Solving for constants.

From eq. (3) and (4), we have:

\(\begin{align}c &= - 12a\\b &= 0\end{align}\)

And from eq. (1) and (2), we get:

\(d = 3\)

Using these values, we get:

\(\begin{align}a &= \frac{3}{{16}}\\c &= - \frac{9}{4}\end{align}\)

So, the cubic polynomial obtained is \(y = \frac{3}{{16}}{x^3} - \frac{9}{4}x + 3\).

Hence, this is the required answer.