SOLUTION: find the vertex, any intercepts &the symmetric line? 1. f(x)=x^2+10x+9

Question 680803: find the vertex, any intercepts &the symmetric line?
1. f(x)=x^2+10x+9
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
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.
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