Question 680803
f(x)=x^2+10x+9

y= x^2+10x+9

x coordinate of vertex = -b/2a= -10/2= -5

plug x= 5 in the equation

x^2+10x+9=y
(-5)^2+10(-5)+9=y
y=-16

Vertex(-5,-16)

The x intercept occurs when y=0, make the equation = 0 and solve for x
x^2 +10x+9 = 0
Factor
(x+9)(x+1) = 0
x = -9 and x = -1, these are the x intercepts
:
y intercept occurs when x=0, substitute 0 for x in the equation, and find y
y = 0^2 - 10(0) + 9
y = 9; :
Axis of symmetry  x = -b/2a= -10/2=-5


x=-5 is the axis of symmetry.