Hi
f(x) = x^2 - 7x - 8
finding x-intercepts when f(x) = 0
x^2 - 7x - 8 = 0
factoring
(x-8)(x+1)=0 Note:SUM of the inner product(-8x) and the outer product(x) = -7x
(x-8)= 0 x = 8 | x-intercept Pt(8,0)
(x+1)=0 x = -1 | x-intercept Pt(-1,0)
finding y-intercepts when x = 0
f(0) = x^2 - 7x - 8 = -8 |y-intercept Pt(0,-8)
finding the vertex by completing the square:
the vertex form of a parabola,
where(h,k) is the vertex
f(x) = x^2 - 7x - 8
f(x) = 1*(x - 7/2)^2 - 49/4 - 8
f(x) = (x - 7/2)^2 - 81/4 Vertex is Pt(7/2,-81/4) Line of symmetry x = 7/2
a = 1 > 0 Parabola opens upward
