How do you find the axis of symmetry, vertex and x intercepts for #y=-x^2-6x-5#?

1 Answer
Oct 2, 2017

#"see explanation"#

Explanation:

#"given a quadratic equation in standard form"#

#•color(white)(x)y=ax^2+bx+c color(white)(x)a!=0#

#"then the x-coordinate of the vertex which is also"#
#"the axis of symmetry is"#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#y=-x^2-6x-5" is in standard form"#

#"with "a=-1,b=-6,c=-5#

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

#"substitute this value into y for y-coordinate"#

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

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

#color(blue)"axis of symmetry is "x=-3#

#"to find the x-intercepts let y = 0"#

#rArr-x^2-6x-5=0#

#"multiply through by "-1#

#rArrx^2+6x+5=0#

#"the factors of + 5 which sum to + 6 are + 5 and + 1"#

#rArr(x+5)(x+1)=0#

#"equate each factor to zero and solve for x"#

#x+5=0rArrx=-5#

#x+1=0rArrx=-1#

#rArrx=-5,x=-1larrcolor(red)" x-intercepts"#
graph{(y+x^2+6x+5)(y-1000x-3000)((x+3)^2+(y-4)^2-0.04)=0 [-10, 10, -5, 5]}