Review
The graph of a quadratic function of the form
f(x) = a x ^{2} + b x + c
is a parabola with vertex at the point (h , k).
where h =  b / 2a
and
and k = f(h) = c  b^{ 2} / 4a
Also the x intercepts (if any) of the graph of f are given by
x1 = [  b + sqrt (b^{ 2}  4 a c) ] / 2 a
and
x2 = [  b  sqrt (b^{ 2}  4 a c) ] / 2 a
It can be easily shown that h = (x1 + x2) / 2
If the graph of f has x intercepts at x1 and x2 then x1 and x2 are solutions to the equation f(x) = 0 so that f may be given by
f(x) = a (x  x1) (x  x2)
Problem 1 : Find the quadratic function whose graph, which is a parabola, has x intercepts at x =  4 and x = 6, and its highest point has a y coordinate equal to 5.
Solution to Problem 1:
The x intercepts are given as  4 and 6 hence f may be written as
f(x) = a (x  (4)) (x  6)
The highest point is the vertex. If x1 and x2 are the x intrcepts of the graph then the x coordinate h of the vertex is given by
h = (x1 + x2) / 2 = ( 4 + 6) / 2 = 1
We now know the x (h = 1) and y coordinates (k = 6) of the vertex which is a point on the graph of the parabola. Hence
f( 1 ) = 6
When f( 1 ) = 6 is substituted in f(x) = a (x  (4)) (x  6), it gives
a (1 + 4)(1  6) = 6
Solve for a to find
a =  6 / 25
The equation of the parabola is given by
f(x) = (  6 / 25) (x + 4) (x  6)
Problem 2 : A parabola, with verical axis, has a vertex at (1 ,  8) and passes through the point (2 ,  6). Find the x intercepts of this parabola.
Solution to Problem 2:
If h and k are the coordinates of the vertex, the quadratic function in vertex form corresponding to this parabola is given by
f(x) = a (x  h)^{ 2} + k
= a (x  1)^{ 2}  8
Use the fact that (2 ,  6) is a point on the graph of the parabola to find coefficient a
 6 = a (2  1)^{ 2}  8
Solve for a to obtain
a = 2
The equation of the parabola is given by
f(x) = 2 (x  1)^{ 2}  8
The x intercepts of the parabola are found by solving
f(x) = 0
Hence
2 (x  1)^{ 2}  8 = 0
Solve for x
2 (x  1)^{ 2}  8 = 0
2 (x  1)^{ 2} = 8
(x  1)^{ 2} = 4
solutions to above equation are
x = 3 and x = 1
The x intercepts are at the points : (3 , 0) and (1 , 0).
Problem 3 : A quadratic function has a minimum value of  2 and its graph has y intercept at (0 , 6) and x intercept at (3 , 0). Find a possible equation for this quadratic function.
Solution to Problem 2:
The minimum value of a quadratic functionic function is equal to k in the quadratic function in vertex form which is given by
f(x) = a (x  h)^{ 2} + k
= a (x  h)^{ 2}  2
We now use the y intercept (0 , 6) given above
6 = a (0  h)^{ 2}  2 = a h^{ 2}  2
Solve the above equation for a
a = 8 / h^{ 2}
Use the x intercept (3 , 0) given in the problem above
0 = a (3  h)^{ 2}  2
Substitute a = 8 / h^{ 2} into the above equation to obtain
0 = ( 8 / h^{ 2} )(3  h)^{ 2}  2
Eliminate the denominators to obtain a quadratic equation in h
h^{ 2}  8 h + 12 = 0
Solve the above equation for h
h = 6 and h = 2.
Use a = 8 / h^{ 2} to find the corresponding values of a
when h = 6 , a = 2 / 9
and when h = 2 , a = 2
The two quadratic functions that are solutions are given by
f(x) = 2 (x  2)^{ 2}  2
f(x) = (2 / 9) (x  6)^{ 2}  2
The graphs (parabolas) of the two quadratic functions obtained are shown below. Check that they have the same x , y intrcepts and minimum value.
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