vertex, axis of symmetry and x-intercepts?
Identify the vertex, axis of symmetry and x-intercepts. f(x)= -x^2-4x+1
thanks for all the help
- DWReadLv 77 years agoBest Answer
y = -x² - 4x + 1
The coefficient of the x² term is negative, so the parabola opens downwards.
The vertex-form equation of a down-opening parabola has the form
y = a(x - h)² + k
a < 0
(h, k) is the vertex
x = h is the axis of symmetry
Convert y = -x² - 4x + 1 to vertex form:
factor out leading coefficient
y = -1(x² + 4x) + 1
complete the square
y = -1(x² + 4x + 2²) + 2² + 1
y = -1(x + 2)² + 5
Now you have all the information you need.
- WedisingheLv 77 years ago
(-2,5) , x=-2 , -2 +/- sqrt(5)