Given that the graph of \(f(x)\) passes through the points \((1,5)\), \((5,6)\), and \((6,2)\), we can use these points to determine the equation of the curve. A common choice for fitting a curve through these points is a quadratic equation of the form \(f(x) = ax^2 + bx + c\).
Let's use the points to set up a system of equations:
1. When \(x = 1\), \(f(1) = a(1)^2 + b(1) + c = a + b + c = 5\).
2. When \(x = 5\), \(f(5) = a(5)^2 + b(5) + c = 25a + 5b + c = 6\).
3. When \(x = 6\), \(f(6) = a(6)^2 + b(6) + c = 36a + 6b + c = 2\).
Now we have a system of three equations with three variables \(a\), \(b\), and \(c\):
\[
\begin{align*}
a + b + c &= 5 \\
25a + 5b + c &= 6 \\
36a + 6b + c &= 2
\end{align*}
\]
Subtracting the first equation from the second equation, we get \(24a + 4b = 1\). Subtracting the first equation from the third equation, we get \(35a + 5b = -3\).
Dividing the equation \(24a + 4b = 1\) by 4, we have \(6a + b = \frac{1}{4}\). Subtracting this from the equation \(35a + 5b = -3\), we get \(29a = -\frac{13}{4}\), which implies \(a = -\frac{13}{116}\).
Substituting this value of \(a\) back into the equation \(6a + b = \frac{1}{4}\), we can solve for \(b\), which gives \(b = \frac{95}{116}\).
Substituting \(a = -\frac{13}{116}\) and \(b = \frac{95}{116}\) into the equation \(a + b + c = 5\), we can solve for \(c\), which gives \(c = \frac{126}{116}\).
So, the equation of the curve is \(f(x) = -\frac{13}{116}x^2 + \frac{95}{116}x + \frac{126}{116}\).
Now, we can use this equation to find the values of \(f(a)\) and \(f(c)\):
- \(f(a) = -\frac{13}{116}a^2 + \frac{95}{116}a + \frac{126}{116}\)
- \(f(c) = -\frac{13}{116}c^2 + \frac{95}{116}c + \frac{126}{116}\)
And the product \(ab + cd\) would be:
\[ab + cd = \left(-\frac{13}{116}a^2 + \frac{95}{116}a + \frac{126}{116}\right) \cdot \left(-\frac{13}{116}c^2 + \frac{95}{116}c + \frac{126}{116}\right)\]
After calculating this expression, you'll get the numerical value of \(ab + cd\).
