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

Question

Kennedy Adams · Accepted Answer

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

Hazel Anglin · Answer

undefined To write a quadratic function given those characteristics, you can use the standard form of a quadratic equation: $ y = ax^2 + bx + c $. To find the specific values of $ a $, $ b $, and $ c $, substitute the given points into the equation and solve the resulting system of equations.