# 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

### 2 Answers

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- 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

where

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.

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