y= -3x^2 - 12x + 11
A.) Identify the coefficients (a, b, and c)
B.) Tell whether the graph opens up or opens down
C.) Find the vertex. Write as a coordinate.
D.) Find the axis of symmetry. Write as an equation.
E.) Find the y-intercept Write as a coordinate.
we have a quadratic function
In this problem
a) we have
a=-3
b=-12
c=11
b) the vertical parabola open downward, because the coefficient a is negative
so
the vertex represent a maximum
c) Convert the given equation in vertex form
y=a(x-h)^2+k
where
(h,k) is the vertex
so
Complete the square
y= -3x^2 - 12x + 11
Factor -3
y=-3(x^2+4x)+11
y=-3(x^2+4x+4)+12+11
y=-3(x+2)^2+23
therefore
the vertex is the point (-2,23)
d) Axis of symmetry
the axis of symmetry for a vertical parabola is equal to the x-coordinate of the vertex
the x-coordinate of the vertex is -2
therefore
the equation of the axis of symmetry is x=-2
e) Find the y-intercept
the y-intercept is the value of y when the value of x is equal to zero
so
For x=0
y= -3(0)^2 - 12(0) + 11
y=11
therefore
the y-intercept is (0,11)