How do you write a quadratic function whose graph has the given characteristics: passes through (5,2), (0,2), (8,-6)?

1 Answer
Nov 23, 2016

#f(x)=-1/3x+5/3x+2#

Explanation:

Let the function be #f(x)#. From the #(0,2)# and #(5,2)#, we can determine that #f(x)-2=a(x-5)(x)# where a is the leading coefficient of the first term, because if you were to shift the function down 2, it would have zeros at 0 and 5. a is negative, given that the position of the third point #(8,-6)#.
The a value is less than one.

#f(x)-2=a(x-5)(x)#
#f(x)=a(x^2-5x)+2#
#f(x)=ax^2-5ax+2#

We know #(8,-6)# is a point on #f(x)#, so plug in the point into the equation, and get a.
#-6=a(8)^2-5a(8)+2#
#-8=64a-40a#
#-8=24a#
#a=-1/3#

Now, write the equation using the fact that #a=-1/3#.
#f(x)=-1/3x^2-5(-1/3)x+2#
#f(x)=-1/3x+5/3x+2#