Find Quadratic Functions given their graphs

Find a quadratic function given its Graph. Examples with detailed solutions are presented. A tutorial with examples on graph of quadratic functions might help in understanding the present examples on finding quadratic equations.

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Review of Quadratic Functions

The general form of a quadratic function is written as
f(x) = a x 2 + b x + c

A quadratic function f in vertex(or standard) form is written as
f(x) = a (x - h) 2 + k
where h and k are the x and y coordinates respectively of the vertex (minimum or maximum) point of the graph.
The graph of of f is a parabola with the vertical line x = h as an axis of symmetry.
The relationship between h and k are
h = - b / 2a and k = f(h) = c - b 2 / (4 a)


Find Quadratic Function Knowing its Vertex and a Point

Example 1

Find the quadratic function f whose graph is shown below.

graph of quadratic function f
Figure 1. Quadratic function f

Solution to Example 1 Let h and k be the coordinates of the the vertex of the graph of function f. From the graph, the vertex (minimum point) is identified as (h , k) = (0 , 2) hence the vertex form of function f may be written as
f(x) = a (x - h)
2 + k = a (x - 0) 2 + 2 = a x 2 + 2
The point (1,3) on the graph of f will now be used to find coefficient a.
f(1) = a (1)
2 + 2 = 3
Solve the above for a to obtain
a = 1
Hence
f(x) = x
2 + 2


Example 2

Find the quadratic function g whose graph is shown below and evaluate g(-3).

graph of quadratic function g
Figure 2. Quadratic function g
Solution to Example 2 The vertex of the graph of function g is a maximum point located at (h , k) = (0 , -1). Hence function g in vertex form is written as
g(x) = a (x - h)
2 + k = a(x - 0) 2 - 1 = a x 2 - 1
Coefficient a will now be found using the point (1,-2) that is on the graph of g.
g(1) = a (1)
2 - 1 = - 2
Solve the above for a to obtain
a = - 1
Hence g(x) is given by
g(x) = - x
2 - 1
g(- 3) = - (- 3)
2 - 1 = - 10


Example 3

Find the quadratic function l whose graph is shown below and calculate the x-intercepts of the graph.

graph of quadratic function l

Figure 3. Quadratic function l

Solution to Example 3 The graph of function l has a vertex (maximum point) located at (h , k) = (2 , 1). Function l in vertex form is written as
l(x) = a (x - h)
2 + k = a (x - 2) 2 + 1
We use the y intercept (0,-7) of the graph of l to find coefficient a as follows.
l(0) = a (0 - 2)
2 + 1 = - 7
Solve the above for a to obtain
a = - 2
Function l(x) is given by
l(x) = - 2 (x - 2)
2 + 1
We now calculate the x intercepts by solving the equation
- 2 (x - 2)
2 + 1 = 0
2 (x - 2)
2 = 1
Extract the square root to obtain the 2 solutions
x = 2 - √(1/2) and x = 2 + √(1/2)
and therefore the x intercepts are located at the points
(2 - √(1/2) , 0) and (2 + √(1/2) , 0)


Find Quadratic Function Knowing its x and y Intercepts

Example 4

Find the quadratic function s in standard form whose graph is shown below.

graph of quadratic function s
Figure 4. Quadratic function s
Solution to Example 4 The graph of function s has two x intercepts: (-1 , 0) and (2 , 0) which means that the equation s(x) = 0 has two solutions x = - 1 and x = 2. Hence s(x) can be written as the product of two factors as follows:
s(x) = a (x + 1)(x - 2)
We now use the y intercept (0,- 4) of the graph of k to find coefficient a as follows.
s(0) = a (0 + 1)(0 - 2) = - 4
Solve the above equation for a to obtain
a = 2
Function s(x) is given by
s(x) = 2 (x + 1) (x - 2)
We expand and simplify to write s(x) in standard form.
s(x) = 2 x
2 - 2 x - 4


Find Quadratic Function Knowing its Axis and Two Points

Example 5

Find the quadratic function m in standard form whose graph is a parabola with an axis of symmetry given by the vertical line x = -3 as shown below.

graph of quadratic function m
Figure 5. Quadratic function m
Solution to Example 5 The graph has an axis of symmetry given by the vertical line x = - 3 hence the x coordinate h of the vertex is equal to - 3 and m(x) may be written as
m(x) = a (x + 3)
2 + k
We now have two unknown a and k to determine. We use the points (-5 , 0) and (-2 , -3/2) shown on the graph of m to write two equations in a and k
The point (-5 , 0) means m(-5) = 0 which gives the equation a(- 5 + 3)
2 + k = 0
The (-2 , -3/2) means m(-2) = -3/2 which gives the equation a(- 2 + 3)
2 + k = -3/2
Simplify to obtain the system of equations
4 a + k = 0
a + k = - 3/2
Solve the system to obtain
a = 1/2 and k = -2
m(x) = (1/2) (x + 3)
2 - 2
Expand and rewrite m(x) in standard form
m(x) = (1/2) x
2 + 3x + 5/2


Find Quadratic Function Knowing Three Points

Example 6

Find the quadratic function w in standard form whose graph is a parabola shown below.

graph of quadratic function w
Figure 6. Quadratic function w
Solution to Example 6 The quadratic function w(x) in standard form is written as follows
w(x) = a x
2 + b x + c
We need to find the coefficients a, b and c. We use the three points on the graph of w to write 3 equations in a, b and c as follows:
point (0,-1/6) gives the equation: w(0) = a (0)
2 + b (0) + c = - 1/6 (eq 1)
point (1 , 0) gives the equation: w(1) = a (1)
2 + b (1) + c = 0 (eq 2)
point (3 , 10/3) gives the equation: w(3) = a (3)
2 + b (3) + c = 10/3 (eq 3)
eq 1 simplifies to
c = - 1/6
Substitute c by - 1/6 in eq 2 and 3 to obtain two equations in a and b
a + b = 1/6
9a + 3 b = 7/2
Solve the above system of equations to obtain
a = 1/2 , b = -1/3
Substitute a, b and c by their values to write the quadratic function w(x) in standard form as follows
w(x) = (1/2) x
2 - (1/3) x - 1/6


More References and links to quadratic functions and parabolas

Graph
Vertex and Intercepts Parabola Problems.
Find the Points of Intersection of a Parabola with a Line.
Parabola Problem with Solution.
Find the Points of Intersection of a Parabola with a Line.
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