Question

How do you find the vertex and the intercepts for `f(x)= -x^2-6x-5`?

Answer 1

see explanation.

Explanation:

`"for the standard quadratic function " y=ax^2+bx+c`

`"the x-coordinate of the vertex " x_(color(red)"vertex")=-b/(2a)`

`"for the function " y=-x^2-6x-5`

`"then " a=-1,b=-6" and " c=-5`

`rArrx_(color(red)"vertex")=-(-6)/(-2)=-3`

`"substitute this value into the function and evaluate for y"`

`rArry_(color(red)"vertex")=-(-3)^2-6(-3)-5=4`

`rArrcolor(magenta)"vertex " =(-3,4)`

`color(blue)"to find intercepts"`

`• " let x=0, in function, for y-intercept"`

`• " let y=0, in function, for x-intercepts"`

`x=0toy=-5larrcolor(red)" y-intercept"`

`y=0to-x^2-6x-5=0`

`rArrx^2+6x+5=0larr" multiply through by - 1"`

`(x+5)(x+1)=0`

`rArrx=-5,x=-1larrcolor(red)" x-intercepts"`
graph{-x^2-6x-5 [-10, 10, -5, 5]}