# Function f(x)=-x2-6x-2 Determine vertex List axis of symmetry Calculate y-intercept Use axis of symmetry and y-intercept find additional point on graph Graph functionNeed correct answer asap

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Given `f(x)=-x^2-6x-2`

(1) If a quadratic function is given in standard form `f(x)=ax^2+bx+c` , then the line of symmetry is `x=-b/(2a)`

**Thus the line of symmetry is `x=6/(2*(-1))=>x=-3` **

(2) The vertex lies on the axis of symmetry, so we find the value of the function at `x=-3` :

`f(-3)=-(-3)^2-6(-3)-2=-9+18-2=7`

**Then the vertex is at `(-3,7)` **

(3) The y-intercept is found by setting x=0, so **the y-intercept is -2.**

(4) Since 0 is three units from the axis of symmetry, the point 3 units on the other side of the axis has the same y-value; thus at x=-6 the value of the function is also -2. **So another point on the graph is (-3,-2).**

(5) The graph: This is a parabola, opening down