
Review Vertex and Intercepts of a Quadratic Functions
The graph of a quadratic function of the form
f(x) = a x ^{2} + b x + c
is a vertical parabola with axis of symmetry parallel to the y axis and has a vertex V with coordinates (h , k), x  intercepts when they exist and a y  intercept as shown below in the graph. When the coefficient a is positive the vertex is the lowest point in the parabola that opens upward and when it is negative, the vertex is the highest point in the parabola that opens downward.
The vertex has coordinates ( h , k) where h =  b / 2a
and k = f(h) = c  b^{ 2} / 4a
The quadratic function of the parabola whose axis is vertical and whose vertex is at the point ( h , k ) is given by
f(x) = a (x  h)^{ 2} + k
The axis of symmetry of the parabola is a vertical line given by the equation: x = h.
The x intercepts (if any) of the graph are the point of intersection of the graph of f and the x axis and are therefore the solutions to the equation a x ^{2} + b x + c = 0 whose solutions are given by the quadratic formulas as follows:
x_{1} = (  b + √(b^{ 2}  4 a c) ) / 2 a
and
x_{2} = (  b  √(b^{ 2}  4 a c) ) / 2 a
It can be easily shown that h = (x_{1} + x_{2}) / 2
If the graph of f has x intercepts at x_{1} and x_{2} then x_{1} and x_{2} are solutions to the equation f(x) = 0 so that f may be given by
f(x) = a (x  x1) (x  x2)
Problems on Vertex and Interecpts of Quadratic Functions
Problem 1
Find the vertex, the x and y intercepts and the axis of symmetry of the graph of the quadratic equation given by f(x) =  x ^{2}  2 x + 3
Solution to Problem 1
There are two methods to find the vertex of a quadratic function. The first one uses the formulae for the coordinates h and k given above and the second method is based on completing the square. We shall use the formulas for h and k given above. We first identify the coefficients a, b and c.
a =  1
b =  2
c = 3
h =  b / 2a =  (2) / (2 × (1)) =  1
k = f(h) = f(1) =  (1) ^{2}  2 (1) + 3 = 4
The vertex is at (1,4).
The x intercepts, if any, are found by solving the equation f(x) = 0. Hence
f(x) =  x ^{2}  2 x + 3 = 0
Factor and solve
x ^{2}  2 x + 3 = (x  1)(x + 3) = 0
gives two solutions: x = 1 and x =  3 ; hence x intercepts are at the points: (1 , 0) and (3 , 0).
The y  intercept is given by f(0)
f(0) =  (0) ^{2}  2 (0) + 3 = 3
Hence the y intercept is at the point (0 , 3).
The axis of symmetry of the graph of the given quadratic function is a vertical line given by x = h.
x =  1
The graph of f(x) =  x ^{2}  2 x + 3 is shown below with the vertex, the x and y intercepts and the axis of symmetry.
Problem 2
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 6.
Solution to Problem 2
The x intercepts are the solutions to the equation f(x) = 0; hence given the x intercepts as  4 and 6, f may be written as a product of factors as follows
f(x) = a (x  (4)) (x  6)
The highest point is the vertex. If x1 and x2 are the x intercepts of the graph then the x coordinate h of the vertex is given by (see formula above)
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)
The graph of the quadratic function f found above is shown below and we can easily check that the x intercepts are at x =  4 and x = 6 and its highest point has a y coordinate equal to 6 as given in the problem above.
Problem 3
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 3
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
Since (2 ,  6) is a point on the graph of the parabola then f(2) =  6, hence
 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 4
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 4
The minimum value of a quadratic 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 intercepts and minimum value.
More References and Links
How to Find x and y Intercepts Of Graphs?
Find Vertex and Intercepts of Quadratic Functions  Calculator
Completing the Square of Quadratic Expressions.
Equation of Parabola
Online Tutorials on Functions and Algebra.
