AnswersNY-Algebra I Term 2 HillcrestSolving Linear-Quadratic Systems

Solving Linear-Quadratic Systems Answers

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3

The first step in determining the solution to the system of equations, y = –x2 – 4x – 3 and y = 2x + 5, algebraically is to set the two equations equal as –x2 – 4x – 3 = 2x + 5. What is the next step?

A
Set y = 0 in y = –x2 – 4x – 3.
B
Factor each side of the equation.
C
Use substitution to create a one-variable equation.
D
Combine like terms onto one side of the equation.
4

Which parabola will have one real solution with the line y = x – 5?

Question illustration
A
y = x2 + x – 4
B
y = x2 + 2x – 1
C
y = x2 + 6x + 9
D
y = x2 + 7x + 4
5

Which line will have no solution with the parabola y – x + 2 = x2?

Question illustration
A
y = –3x –3
B
y = –2x –3
C
y = 2x –3
D
y = 3x –3
6

Which represents the solution(s) of the graphed system of equations, y = x2 + 2x – 3 and y = x – 1?

Question illustration
A
(1, 0) and (0, –1)
B
(–2, –3) and (1, 0)
C
(0, –3) and (1, 0)
D
(–3, –2) and (0, 1)

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