y=x^2+4x-6

axis, vertex, roots
in Trigonometry Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

y=x^2+4x-6
axis, vertex, roots

The roots are found by setting y = 0 and completing the square.

x^2 + 4x - 6 = 0
x^2 + 4x = 6
x^2 + 4x + 2^2 = 6 + 2^2       add the square of (1/2) * b, the x coefficient
x^2 + 4x + 2^2 = 6 + 4 + 10
(x + 2) * (x + 2) = 10
x + 2 = ±√10
x + 2 = ±3.16228
x = ±3.16228 - 2
x = 3.16228 - 2        and   x = -3.16228 - 2
x = 1.6228               and   x = -5.16228

The roots, where the plot crosses the x axis, are (-5.16228, 0) and (1.6228, 0)

For the vertex, x = (-b/2a), and y = -(b^2 - 4ac)/4a

x = (-4/(2*1))
x = -4/2
x = -2

y = -(b^2 - 4ac)/4a
y = -(4^2 - 4(1)(-6))/4(1)
y = -(16 + 24)/4
y = -40/4
y = -10

The vertex is (-2, -10)

The axis is the vertical line x = -2
by Level 11 User (78.4k points)

No related questions found

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,516 questions
100,285 answers
2,420 comments
737,296 users