What is the axis of symmetry and vertex for the graph #y=x^2-10x+2#?
1 Answer
Mar 21, 2016
vertex = (5,-23) , x = 5
Explanation:
The standard form of a quadratic is y
# = ax^2+bx+c # The function :
# y = x^2-10x+2 " is in this form " # with a = 1 , b = -10 and c = 2
the x-coord of vertex
#= -b/(2a) = -(-10)/2 = 5 # now substitute x = 5 into equation to obtain y-coord
y-coord of vertex
# = (5)^2 - 10(5) + 2 = 25-50+2 = -23# thus vertex =( 5 , -23)
The axis of symmetry passes through the vertex and is parallel to the y-axis with equation x = 5
Here is the graph of the function with the axis of symmetry.
graph{(y-x^2+10x-2)(0.001y-x+5)=0 [-50.63, 50.6, -25.3, 25.32]}
