Find the axis of symmetry, vertex and y-intercept of the function
f(x) = -x² - 2x – 5

The correct answer is: 0

This quadratic function is in standard form, f(x)=ax²+bx+c.
For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.
In f(x)= -x² - 2x – 5, a= -1, b= -2, and c= -5. So, the equation for the axis of symmetry is given by

x = −(-2)/2(-1)

x = 2/-2

x = -1
The equation of the axis of symmetry for f(x)= -x² - 2x – 5 is x = -1.
The x coordinate of the vertex is the same:

h = -1
The y coordinate of the vertex is :

k = f(h)

k = -(h)² - 2(h) - 5

k = -(-1)² - 2(-1) - 5

k = -1 + 2 - 5

k = -4
Therefore, the vertex is (-1 , -4)
For finding the y- intercept we firstly rewrite the equation by substituting 0 for x.

y = -(0)² - 2(0) - 5

y = 0 - 0 - 5

y = -5
Therefore, Axis of symmetry is x = -1
Vertex is ( -1 , -4)
Y- intercept is -5.