Question

What is the vertex of the graph of the function below?y = x^2 + 10x + 24O A. (-4,-1)O B. (-5, -1)O C. (-5,0)O D. (4,0)

Answer 1

For any given parabola in the form

`f(x)=ax^2+bx+c`

The vertex is the point:

`V=(-(b)/(2a),f(-(b)/(2a)))`

This way,

`\begin{gathered} y=f(x)=x^2+10x+24 \\ \rightarrow a=1 \\ \rightarrow b=10 \\ \rightarrow c=24 \\ \\ \rightarrow-(b)/(2a)=-(10)/(2\cdot1)=-5 \\ \\ \rightarrow f(-5)=(-5)^2+10(-5)+24=-1 \end{gathered}`

Therefore, the vertex is:

`V(-5,-1)`

Answer: Option B