Lesson 8: Systems of Linear and Quadratic Equations
Check my work
1. Solve the system of equations.
y = 2x^2 - 3
y = 3x - 1
a. no solution
b. (-1/2, 5), (2, -5/2)
c. (-1/2, -5/2), (2,5)***
d. (1/2, 5/2), (2, 5)
2.how many real number solutions does the equation have 0 = -3x^2 + x - 4
a. 0***
b. 1
c. 2
d. 3
3. solve the equation by completing the square. If necessary round to the nearest hundredth.
x^2 - 18x = 19
a. 1; 19
b. -1; 19***
c. 3; 6
d. -3; 1
4. solve. x^2 - 81 = 0
a. 0
b. -9
c. -9, 9***
d. 9
5. which model is most appropriate for the data shown in the graph below? (need wedsite to know the problem)
a. quadratic
b. linear
c. exponential***
d. line
for questions 6-9, match the equation to its corresponding graph.
(need wedsite to know the problem)
a, b, c, d
6. a y=-2x^2+2
7. d y=-x^2
8. c y=2x^2
9. b y=3x^2-4
Graph the quadratic functions y = -2x^2 and y = -2x^2 + 4 on a separate piece of paper. Using those graphs, compare and contrast the shape and position of the graphs.
y=-2x^2+4 is just 4 units up from y=-2x^2
19 answers
#5, can't tell , can't see any graph
yes
1. C (-1/2, -5/2), (2,5)
2. A 0
3. B -1;19
4. C -9,9
5. C Exponential
MATCHING
6. A
7. D
8. C
9. B
10. ESSAY
This answer is correct. The two parabolas have the same shape and axis of symmetry, but the second parabola has been vertically translated upwards by 4 units, resulting in a vertex at (0, 4) instead of (0, 0). The second parabola also has a maximum point since it opens downwards, and this maximum point occurs at the vertex.