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



 

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