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sp2606 [1]
3 years ago
8

Which represents the solution(s) of the graphed system of equations, y = x2 + x – 2 and y = 2x – 2?

Mathematics
2 answers:
Assoli18 [71]3 years ago
6 0
Your answer is B.

Plug in the values for x and y in the equations. The only points which make the system of equations true is (0, -2) & (1,0).
Alisiya [41]3 years ago
4 0

Answer: the second choice: (0, - 2) and (1, 0).


Explanation:


1) The solution of a graphed system of equations is the points of intersection of the equations.


2) Hence, this method requires to graph each function in one Cartesian coordinate system.


3) To graph the function y = x² + x  - 2, you can follow these steps

  • draw the perpendicular x(horizontal)-axis, and y(vertical)- axis
  • mark the divisions in each axis (1 unit is a good scale for this case)
  • notice that y = x² + x  - 2 is a parabola
  • write y = x² + x  - 2 in its vertex form following these steps:

       y = (x² + x)  - 2

       y + 1/4 =  (x² + x + 1/4)  - 2

        y + 1/4 = (x + 1/2)²  - 2

        y = (x + 1/2)²  - 2 - 1/4

        y = (x + 1/2)² - 9/4

        Compare with the vertex form y = (x + h)² + k, where the coordinates of the vertex are (h,k). Therefore, the vertex is (1/2, - 9/4).

  • The parabola open upwards (because the coefficient of the quadratic term is positive).
  • Find the y-intercept (x =0) and x-intercepts (y = 0), and write a table with some values:
  • x       y = (x + 1/2)² - 9/4
  • -2            0
  • -1           - 2
  • -1/2       -2.25
  • 0         - 2
  • 1/2         -1.25
  • 1             0
  • 2           4

4) To graph the function y = 2x - 2, you can follow these steps:

  • Use the same Cartesian coordinate system
  • Notice it is a line
  • The slope is 2 (the coefficient of x)
  • The y-intercept is - 2( the constant term)
  • Build a table (use the same x-values used for the first graph)
  • x        y = 2x - 2
  • -2          - 6
  • -1           - 4
  • -1/2       - 3
  • 0          - 2
  • 1/2        -  1
  • 1             0
  • 2            2

The tables show that these points are common to the two functions: (0, - 2) and (1,0).


The graphs are shown in the figure attached.


You can see that the parabola (blue curve) and the line (purple line) intercept each other at those two points ( 0, - 2) and (1,0).

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