Question

Which of the following is true for the graph of the quadratic function? f(x)=1/5(x-3)^2-5

Answer 1

Given function f(x) = 1/5(x-3)^2-5

[we know that (a-b)^2 = a^2+b^2-2ab]

y = 1/5(x^2+9-6x)-5

y = (1/5)x^2-(6/5)x+9/5-5

y = (1/5)x^2-(6/5)x-16/5 ---> (1)

Compare it quadratic form ax^2+bx+c = 0

a = 1/5, b = -6/5 , c = -16/5

Axis of symmetry x = -b/2a = -(-6/5)/2(1/5) = 6/2 = 3

To find vertex substitute the x value in (1).

y = (1/5)9-(6/5)(3)-16/5

y = 9/5-18/5-16/5

y = -25/5

y = -5

Vertex of parabola (3,-5).

Graph of f(x) = 1/5(x-3)^2-5