Asked by smc
If f(x)=2(x-3)^2-5 find the vertex, the x intercepts the y intercepts sketch the graph, choose two points on the graph and find the rate of change
Answers
Answered by
Henry
F(x) = Y = 2(x-3)^22(-5
Y = 2(x^2--6x+9)-5
Y = 2x^2-12x+18-5
Y = 2x^2-12x+13
h = -B/2A = 12/4 = 3
k = 2(3-3)^2-5 = -5
V(h,k) = (3,-5)
Use Quad. Formula and get:
X = 4.5,and 1.5. = X- Intercepts
To find Y-int.,replace x in given Eq
with 0 and solve for y.
Y-int = 4.
Y = 2(x^2--6x+9)-5
Y = 2x^2-12x+18-5
Y = 2x^2-12x+13
h = -B/2A = 12/4 = 3
k = 2(3-3)^2-5 = -5
V(h,k) = (3,-5)
Use Quad. Formula and get:
X = 4.5,and 1.5. = X- Intercepts
To find Y-int.,replace x in given Eq
with 0 and solve for y.
Y-int = 4.
Answered by
Henry
Correction: F(x) = Y = 2(x-3)^2-5
Use the following points for graphing:
(X,Y)
(1,3)
(2,-7)
V(3,-5)
(4,-3)
(5,3)
Use the following points for graphing:
(X,Y)
(1,3)
(2,-7)
V(3,-5)
(4,-3)
(5,3)
Answered by
Henry
Correction: (2,-7) should be (2,-3).
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