Question

Where is the vertex of the parabola? y = x2 + 6x + 9

a) on the y-axis
b) above the x-axis
3) below the x-axis
4) on the x-axis

Answer 1

`image`

now that point is right on the x-axis, right between the II and III Quadrants.

Answer 2

`f(x)=ax^2+bx+c\\\\\text{The formula of a vertex:}\\(h,\ k)\ \text{where}\ h=(-b)/(2a)\ \text{and}\ k=f(h)=(-(b^2-4ac))/(4a)`

`\text{We have}\\\\f(x)=x^2+6x+9\to a=1,\ b=6,\ c=9`

`\text{Substitute:}\\\\h=(-6)/(2\cdot1)=(-6)/(2)=-3\\\\k=f(-3)=(-3)^2+6(-3)+9=9-18+9=0`

`\text{The vertex}\ (-3,\ 0)\ \text{therefore your answer is}\\\\\boxed{4)\ on\ the\ x-axis}`

`\text{Other method.}\\\\\text{The vertex formula of a quadratic function}\\\\f(x)=a(x-h)^2+k`

`(h,\ k)-vertex`

`\text{We have}\ y=x^2+6x+9=\underbrace{x^2+2\cdot x\cdot3+3^2}_((a+b)^2=a^2+2ab+b^2)=(x+3)^2=(x-(-3))^2+0\\\\\text{therefore}\ h=-3,\ k=0`

`The\ vertex\ is\ in\ (-3,\ 0)-on\ the\ x-axis`