SOLUTION: The quadratic function f(x)=ax^2+bx+c has the following characteristics: (i) passes throught the point (2,4); (ii) has a maximum value of 6 when x=4; and (iii) has a zero of x=4

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The quadratic function f(x)=ax^2+bx+c has the following characteristics: (i) passes throught the point (2,4); (ii) has a maximum value of 6 when x=4; and (iii) has a zero of x=4      Log On


   

Question 981065: The quadratic function f(x)=ax^2+bx+c has the following characteristics:
(i) passes throught the point (2,4); (ii) has a maximum value of 6 when x=4; and (iii) has a zero of x=4+2√3
Find the values of a,b,c

Thanks a bunch!
Answer by josgarithmetic(39614) About Me  (Show Source):
You can put this solution on YOUR website!
Vertex is given: (6,4), so you have a%2A4%5E2%2Bb%2A4%2Bc=6 or 16a%2B4b%2Bc=6.



Using standard form,
y=a%28x-4%29%5E2%2B6
but you do not yet know value of a.
-
The parabola passes through point (2,4).
a%28x-4%29%5E2%2B6=y
a%28x-4%29%5E2=y-6
a=%28y-6%29%2F%28x-4%29%5E2
substitute the coordinates,
a=%284-6%29%2F%282-4%29%5E2
a=-2%2F%28-2%29%5E2
highlight%28a=-1%2F2%29---------one of the values found.

Standard form again:
y=%28-1%2F2%29%28x-4%29%5E2%2B6
%28-1%2F2%29%28x%5E2-8x%2B16%29%2B6
-%281%2F2%29x%5E2%2B4x-8%2B6
-%281%2F2%29x%5E2%2B4x-2----------this is general form.
-
Compare to the general form,
highlight%28system%28b=4%2Cc=-2%29%29


---
The information about the given zero does not play much part in finding the equation for the function:

If z is the other zero, you expect that %28z%2B4%2B2sqrt%283%29%29%2F2=4 because the x value for the vertex must be in the middle of the two zeros.
z=8-4-sqrt%283%29
z=4-sqrt%283%29
-
The two zeros are 4%2Bsqrt%283%29 and 4-sqrt%283%29.
-
y=0=a%28x-%284%2Bsqrt%283%29%29%29%28x-%284-sqrt%283%29%29%29