How do you find the vertex and intercepts for f(x)= -x^2 + 6x + 3?

1 Answer
Jan 8, 2017

The vertex is =(3,6)
The intercepts are (0,3), (3+2sqrt3,0) and (3-2sqrt3,0)

Explanation:

The vertex is the highest point or lowest point of the equation,

we complete the squares

We rewrite f(x) as

f(x)=-x^2+6x+3

=-(x^2-6x)+3

=-(x^2-6x+9)+3+9

=-(x-3)^2+12

We compare this to f(x)=a(x-h)^2+k

and the vertex is (h,k)=(3,6)

The axis of symmetry is x=3

To find the intercepts,

first,

Let x=0, =>, (y=3)

second,

Let y=0, then

-(x-3)^2+12=0

(x-3)^2=12

x-3=+-sqrt12=+-2sqrt3

x=3+-2sqrt3

Therefore,

The points are (3+2sqrt3,0) and (3-2sqrt3,0)

The points are (6.464,0) and (-0.464,0)

graph{(y-(-x^2+6x+3))((x-3)^2+(y-12)^2-0.01)(y-1000x+3000)=0 [-18.1, 17.93, -2.36, 15.67]}