How do you find the vertex and the intercepts for y=x^2+9x+8?

1 Answer
Aug 16, 2016

Vertex is at (-9/2,-49/4); y-intercept is at (0,8); x-intercept at (-1,0) & (-8,0)

Explanation:

y=x^2+9x+8 = (x+9/2)^2-81/4+8=(x+9/2)^2-49/4:.Vertex is at (-9/2,-49/4) Putting x=0 we get y-intercept as y=8;
Putting y=0 we get x-intercept as x^2+9x+8=0 or (x+8)(x+1)=0:. x= -8 ; x=-1 graph{x^2+9x+8 [-40, 40, -20, 20]}[Ans]