How do you find the y-intercept, the axis of symmetry and the vertex of f(x)=-x^2-2x-5

Answer 1

The y-intercept is found by letting x=0 in f(x). f(0) = -5.

The axis of symmetry is found from the quadratic equation [ -b +/- sqrt(b²-4ac) ] / [2a], which gives the roots. the x-coordinate of the vertex lies midway between the roots, so just note the mean of the roots is

-b / [2a] = -(-2) / [2(-1)] = 2 / -2 = -1.

Answer: x = -1 is the axis of symmetry.

Answer 2

There is a formula to find the x coordinate of the vertex.

First, identify the a and b term in the equation.

a=-1

b=-2

c=-5, this one isn't used to find the vertex

x=-b/2a = -(-2)/[(2)(-1)] = 2/-2 = -1

Now that we have the x coordinate, we can plug that into the original equation to find y

y=-(-1)²-2(-1)-5

y=-1+2-5 = -4

Bringing these 2 values together gives the vertex of (-1,-4)

The axis of symmetry comes from the x coordinate of the vertex, which is -1

The axis of symmetry is x=-1

Finally the y intercept comes from setting the x coordinate equal to 0 in the original equation and solving.

y=-(0)²-2(0)-5

y=-5

The y intercept is (0,-5)