SOLUTION: Consider the quadratic function [f(x)=2(x-3)^2+5]
a)Identify the vertex
I believe this is [v=(3,5)]
b)Identify the axis of symmetry (x=?)
c)Identify the y-intercept
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-> SOLUTION: Consider the quadratic function [f(x)=2(x-3)^2+5]
a)Identify the vertex
I believe this is [v=(3,5)]
b)Identify the axis of symmetry (x=?)
c)Identify the y-intercept
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Question 953793: Consider the quadratic function [f(x)=2(x-3)^2+5]
a)Identify the vertex
I believe this is [v=(3,5)]
b)Identify the axis of symmetry (x=?)
c)Identify the y-intercept Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! We are given f(x)=2(x-3)^2+5
expand function
f(x) = 2*(x^2 -6x +9) + 5
f(x) = 2x^2 -12x +18 +5
f(x) = 2x^2 -12x +23
The formula for the axis of symmetry is
x = -b/(2a) = -(-12) / (2*2) = 12/4 = 3
f(3) = 2*9 -12*3 +23
f(3) = 18 -36 +23 = 5
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a) vertex is (3, 5)
b) axis of symmetry is x = 3
c) y intercept is calculated by setting x = 0 for f(x)
f(0) = 2*0^2 -12*0 +23
f(0) = 0 - 0 +23
f(0) = 23
y intercept is (0, 23)
