\begin{tabular}{|c|c|} \hline$x$ & $y$ \\ \hline-1 & -3 \\ \hline 0 & -3 \\ \hline 3 & 33 \\ \hline \end{tabular}
Find a quadratic function to model the values in the table.
Answer (1)
Substitute the given points into the general quadratic equation y = a x 2 + b x + c to form a system of equations.
Solve for c directly from the point (0, -3), which gives c = − 3 .
Substitute c = − 3 into the other equations and simplify to find a and b .
Determine that a = 3 and b = 3 , so the quadratic function is y = 3 x 2 + 3 x − 3 .
Explanation
Understanding the Problem We are given a table of x and y values and asked to find a quadratic function that models these values. A quadratic function has the general form y = a x 2 + b x + c , where a , b , and c are constants. We need to determine the values of these constants using the given data points.
Setting up the Equations We have three data points: ( − 1 , − 3 ) , ( 0 , − 3 ) , and ( 3 , 33 ) . We can substitute these values into the quadratic equation to create a system of three equations with three unknowns:
For ( − 1 , − 3 ) : − 3 = a ( − 1 ) 2 + b ( − 1 ) + c ⇒ − 3 = a − b + c
For ( 0 , − 3 ) : − 3 = a ( 0 ) 2 + b ( 0 ) + c ⇒ − 3 = c
For ( 3 , 33 ) : $33 = a(3)^2 + b(3) + c \Rightarrow 33 = 9a + 3b + c
Simplifying the Equations From the second equation, we directly find that c = − 3 . Now we substitute this value into the first and third equations:
− 3 = a − b − 3 ⇒ a − b = 0
33 = 9 a + 3 b − 3 ⇒ 36 = 9 a + 3 b ⇒ 12 = 3 a + b
Solving for a and b Now we have a system of two equations with two unknowns:
a − b = 0
3 a + b = 12
From the first equation, we have a = b . Substituting this into the second equation, we get:
3 a + a = 12 ⇒ 4 a = 12 ⇒ a = 3
Since a = b , we also have b = 3 .
The Quadratic Function Now we have found the values of a , b , and c : a = 3 , b = 3 , and c = − 3 . Therefore, the quadratic function is:
y = 3 x 2 + 3 x − 3
Final Answer The quadratic function that models the values in the table is y = 3 x 2 + 3 x − 3 .
Examples
Quadratic functions are incredibly useful for modeling various real-world scenarios, such as the trajectory of a ball thrown in the air or the shape of a satellite dish. Imagine you're designing a suspension bridge; the curve of the cables can be modeled using a quadratic function to ensure stability and even weight distribution. By understanding how to derive these functions from data points, you can predict and control outcomes in engineering, physics, and even economics, where they help model cost curves and revenue projections.
