x 0 1 2 y 1 3 2
The Correct Answer and Explanation is:
Correct Answer:
To find the equation that models the relationship between xxx and yyy for the given data:
| xxx | 0 | 1 | 2 |
|---|---|---|---|
| yyy | 1 | 3 | 2 |
We want to find a quadratic function of the form:y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c
Step 1: Use the given points
Substitute the values of xxx and yyy into the general form y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c:
- When x=0,y=1x = 0, y = 1x=0,y=1:
1=a(0)2+b(0)+c⇒c=11 = a(0)^2 + b(0) + c \Rightarrow c = 11=a(0)2+b(0)+c⇒c=1
- When x=1,y=3x = 1, y = 3x=1,y=3:
3=a(1)2+b(1)+1⇒3=a+b+1⇒a+b=23 = a(1)^2 + b(1) + 1 \Rightarrow 3 = a + b + 1 \Rightarrow a + b = 23=a(1)2+b(1)+1⇒3=a+b+1⇒a+b=2
- When x=2,y=2x = 2, y = 2x=2,y=2:
2=a(2)2+b(2)+1⇒2=4a+2b+1⇒4a+2b=12 = a(2)^2 + b(2) + 1 \Rightarrow 2 = 4a + 2b + 1 \Rightarrow 4a + 2b = 12=a(2)2+b(2)+1⇒2=4a+2b+1⇒4a+2b=1
Step 2: Solve the system of equations
From above:
- Equation (1): a+b=2a + b = 2a+b=2
- Equation (2): 4a+2b=14a + 2b = 14a+2b=1
Use substitution or elimination:
Multiply equation (1) by 2:
2a+2b=42a + 2b = 42a+2b=4
Now subtract from equation (2):
(4a+2b)−(2a+2b)=1−4⇒2a=−3⇒a=−32(4a + 2b) – (2a + 2b) = 1 – 4 \Rightarrow 2a = -3 \Rightarrow a = -\frac{3}{2}(4a+2b)−(2a+2b)=1−4⇒2a=−3⇒a=−23
Substitute back:
−32+b=2⇒b=72-\frac{3}{2} + b = 2 \Rightarrow b = \frac{7}{2}−23+b=2⇒b=27
Final Equation:
y=−32×2+72x+1y = -\frac{3}{2}x^2 + \frac{7}{2}x + 1y=−23x2+27x+1
Explanation (Like in Textbooks):
This problem involves identifying the rule that governs the relationship between two variables, xxx and yyy, based on a table of values. Since the values do not change at a constant rate, a linear model is not suitable. The pattern suggests a curved relationship, indicating a quadratic function.
To model this, use the standard quadratic form y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c. By substituting the known values from the table into this form, a system of equations is created. Solving this system gives the values of the coefficients aaa, bbb, and ccc, which define the specific quadratic equation. Substituting these back into the standard form provides the equation that fits all three data points.
This approach demonstrates how equations can be built from data and how quadratic functions model nonlinear patterns in relationships.
