Question

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

Answer 1

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]