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
(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` :
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