Question

How do you find the vertex of a parabola `f(x)=x^2 +10x-11`?

Answer 1

Vertex of the parabola is at :`color(blue)((-5, -36)`

Explanation:

The Standard form for the quadratic equation is :

`y=f(x)=ax^2+bx+c = 0`

Given:

`y=f(x) = x^2+10x-11`

where `a=1; b=10 and c= -11`

Find the x-coordinate of the vertex using the formula `-b/(2a)`

`x=-10/(2*1)`

`rArr x=-5`

To find the y-coordinate value of the vertex, set `x=-15` in `y= x^2+10x-11`

`y=(-5)^2+10(-5)-11`

Simplify the right-hand side.

`rArr 25-50-11`

`rArr 25-61`

`rArr -36`

`y=-36`

Hence,

Vertex of the parabola is at :`color(blue)((-5, -36)`

Also observe that, since the coefficient of `x^2` term `(a)` is less than zero, the parabola opens down and the graph has a minimum at the vertex.

Please refer to the image of the graph below: