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

##### 1 Answer

May 4, 2017

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]}