N 2
we have
f(x)=-x^2-4x+12
this is a vertical parabola open downward (because the leading coefficient is negative)
The vertex represents the maximum
Convert the given equation into vertex form
f(x)=a(x-h)^2+k
where
(h,k) is the vertex
The axis of symmetry is equal to x=h
so
f(x)=-x^2-4x+12
factor -1
f(x)=-(x^2+4x)+12
Complete the square
f(x)=-(x^2+4x+2^2-2^2)+12
f(x)=-(x+2)^2+16
the vertex is (-2,16)
The axis of symmetry is x=-2
Find out the y-intercept (value of f(x) when the value of x=0)
For x=0
f(x)=-(0)^2-4(0)+12
f(x)=12
The y-intercept is (0,12)
Graph the function
we have the points
(-2,16) vertex
(0,12) y-intercept
Find out te x-intercepts (values of x when the value of f(x)=0
-x^2-4x+12=0
solve the quadratic equation using the formula
a=-1
b=-4
c=12
substitute
[tex]x=\frac{-(-4)\pm\sqrt[]{-4^2-4(-1)(12)}}{2(-1)}[/tex][tex]x=\frac{4\pm8}{-2}[/tex]
The values of x are
x=
