The graph of a quadratic function has a y intercept at 0,5 and a minimum at 3,-4 ?

2 Answers
Apr 25, 2018

#f(x) = x^2 - 6x + 5#

Explanation:

#f(x)= ax^2 + bx + c#

#5 = f(0) = a( 0^2) + b(0) + c #

#c = 5#

The minimum #y# is at #x=-b/{2a}.#

#-b/{2a} = 3#

#b = -6a #

#(3,-4)# is on the curve:

#-4 = f(3) = a(3)^2 + (-6a) (3) + 5 #

# -9 = -9 a #

# a = 1#

#b = -6a = -6 #

#f(x) = x^2 - 6x + 5#

Check: #f(0)=5 quad sqrt#

Completing the square,

# f(x) = (x^2 - 6x + 9) -9 + 5 = (x- 3)^2 -4 # so #(3,-4)# is the vertex.#quad sqrt#

Apr 25, 2018

#y=(x-3)^2-4#

Explanation:

Assuming that the equation of such quadratic graph is requested:

#y=a(x-h)^2+k# => Equation of parabola in vertex form where:

#(h, k)# is the vertex, for #a > 0# the parabola opens up which

makes the vertex the minimum, so in this case #(3, -4)# is the

vertex then:

#y=a(x-3)^2-4# => the #y# intercept is at: #(0, 5)#:

#5=a(0-3)^2-4# => solving for #a#:

#5=9a-4#

#9=9a#

#a=1#

Thus the equation of the graph is:

#y=(x-3)^2-4#