Examkin

Question

Difficulty: MediumSystems of Linear and Non-Linear Equations

Consider the system of equations consisting of the quadratic function f(x)=(x−2)2−3f(x) = (x - 2)^2 - 3 and the linear function g(x)=3x−9g(x) = 3x - 9. If the graphs of these functions intersect at two distinct points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), what is the value of x1y2+x2y1x_1 y_2 + x_2 y_1?

  1. A
    -15
  2. B
    -8
  3. -3Answer
  4. D
    15
  5. E
    24

Answer

The correct answer is −3-3. The intersection points of the two functions are (2,−3)(2, -3) and (5,6)(5, 6), and evaluating the expression x1y2+x2y1x_1 y_2 + x_2 y_1 gives −3-3.
To find the points where the graphs of the functions intersect, we set their expressions equal to each other: (x−2)2−3=3x−9(x - 2)^2 - 3 = 3x - 9. Expanding the squared term gives x2−4x+4−3=3x−9x^2 - 4x + 4 - 3 = 3x - 9, which simplifies to x2−4x+1=3x−9x^2 - 4x + 1 = 3x - 9. Moving all terms to the left side yields x2−7x+10=0x^2 - 7x + 10 = 0. Factoring this quadratic equation gives (x−2)(x−5)=0(x - 2)(x - 5) = 0, so the xx-coordinates of the intersection points are 22 and 55. Substituting these back into the linear equation gives the corresponding yy-coordinates: y=3(2)−9=−3y = 3(2) - 9 = -3 and y=3(5)−9=6y = 3(5) - 9 = 6, resulting in the intersection points (2,−3)(2, -3) and (5,6)(5, 6). Finally, evaluating the requested expression gives (2)(6)+(5)(−3)=12−15=−3(2)(6) + (5)(-3) = 12 - 15 = -3.

Step-by-Step Solution

1
Set the quadratic and linear functions equal to find the xx-coordinates of their intersection points.
(x−2)2−3=3x−9(x - 2)^2 - 3 = 3x - 9
At the points of intersection, the values of f(x)f(x) and g(x)g(x) must be equal.
2
Expand the binomial squared term (x−2)2(x - 2)^2.
x2−4x+4−3=3x−9x^2 - 4x + 4 - 3 = 3x - 9
Expanding the binomial is necessary to combine like terms and write the equation in standard quadratic form.
3
Move all terms to one side to set the equation to zero.
x2−7x+10=0x^2 - 7x + 10 = 0
A quadratic equation must be set to zero before factoring or applying the quadratic formula.
4
Factor the quadratic equation.
(x−2)(x−5)=0  ⟹  x1=2 and x2=5(x - 2)(x - 5) = 0 \implies x_1 = 2 \text{ and } x_2 = 5
Factoring determines the xx-coordinates of the intersection points.
5
Substitute the xx-values into the linear equation g(x)=3x−9g(x) = 3x - 9 to find the corresponding yy-coordinates.
y1=3(2)−9=−3y_1 = 3(2) - 9 = -3 y2=3(5)−9=6y_2 = 3(5) - 9 = 6
This yields the two intersection points: (2,−3)(2, -3) and (5,6)(5, 6).
6
Calculate the value of the expression x1y2+x2y1x_1 y_2 + x_2 y_1.
(2)(6)+(5)(−3)=12−15=−3(2)(6) + (5)(-3) = 12 - 15 = -3
This evaluates the requested secondary value using the intersection coordinates.

Key Concept

Solving systems of linear and non-linear equations by substitution and algebraic manipulation.
Rate this question

Topics

Systems of Linear and Non-Linear Equations

More questions from this topic