Graph Quadratic Functions

Graphs of quadratic functions of the vertex form f(x) = a (x - h)2 + k and of the standard form f(x) = a x2 + b x + c are presented with several examples and their detailed solutions.
We start with the graph of the basic quadratic function f(x) = x2, then we graph examples of quadratic functions in vertex form and then in standard form.
A tutorial with several examples on find quadratic functions given their graphs is also included in this website.
Quadratic functions are polynomial functions of degree 2 (x squared) and the word "quadratic" comes from the Latin word "quadratus" which means square.

Graph of the basic quadratic function: f(x) = x2

Example 1

Find points on the graph of the function f(x) = x 2 and graph it.
We first note that
1) because the square of real number is either positive or zero, x
2 ≥ 0 and therefore the graph of f touches the x axis at x = 0 and is above the x axis for all other values of x.
2) f(-x) = (-x)
2 = x 2 = f(x) , hence function f is even and its graph is symmetric with respect to the y axis.
3) The range of values of y = f(x) is : y ≥ 0
Let us a table to find points on the graph of function f
x f(x) = x2
0 02 = 0
- 1 (-1)2 = 1
1 12 = 1
- 2 (-2)2 = 4
2 22 = 4
- 3 (-3)2 = 9
3 32 = 9

We now plot the points in the table and join them to obtain the graph of f(x) = x
2 called the parabola .

graph of basic quadratic function f(x) = x^2

Figure 1. graph of basic quadratic function f(x) = x2

The graph has a
vertex at (0,0) (minimum point) the vertical line x = 0 (y axis) as the axis of symmetry .

Graph of quadratic functions in vertex form: g(x) = a(x - h)2 + k

A quadratic function in vertex form g(x) = a(x - h) 2 + k is the basic quadratic function f(x) = x 2 that has been transformed.
1) From x
2 to (x - h) 2 : shift h units right if h is positive or - h units left if h is negative.
2) From (x - h)
2 to a(x - h) 2 : vertical stretching (|a| > 1) or compression (|a| < 1) plus reflection on x axis if a is negative.
3) From a(x - h)
2 to a(x - h) 2 + k: a vertical shift by k units.
Conclusion
A quadratic function in
vertex form g(x) = a(x - h) 2 + k has a graph
1) with a
vertex (minimum or maximum) at (h , k)
2) A
vertical axis of symmetry x = h.
3) The graph of g which is a
parabola opens upward if the coefficient a is positive and therefore the vertex is a minimum point or downward if a is negative and therefore the vertex is a maximum point .
4) The
range of values of y = g(x) is : y ≥ k if a is positive or y ≤ k if a is negative.

Example 2

a) Identify the vertex of the graph of function g(x) = 4 (x - 1) 2 - 1, its axis of symmetry and decide whether the graph opens upward or downward.
b) Find the points of intersection of graph of g with the axis (x intercepts) and y axis (y intercepts).
c) Use the vertex, the x and y intercepts, some more points on the graph of g, and graph it.
d) What is the range of values of y = g(x)?

Solution to Example 2

a) Comparing g(x) = 4 (x - 1) 2 - 1 to g(x) = a(x - h) 2 + k, h = 1, k = -1 and a = 4.
The vertex is at the point (h , k) = (1 , -1).
The axis of symmetry is the vertical line x = h = 1
a is positive , the parabola (graph of g) opens upward.
b) The y intercept is a point on the y axis therefore x = 0 for this point and its y coordinate given by g(0) = 3. Therefore the y intercept is at (0 , 3)
The x intercept is a point on the x axis therefore y = g(x) = 0 for this point and its x coordinate(s) are found by solving
g(x) = 4 (x - 1)
2 - 1 = 0
4 (x - 1)
2 = 1
2 (x - 1) = ~+mn~ √1
2(x - 1) = 1 gives x = 3/2 = 1.5
2(x - 1) = - 1 gives x = 1/2 = 0.5
Two x intercepts are at: (1/2 , 0) and (3/2 , 0)
c) One more point on the graph of g.
g(2) = 4 (2 - 1)
2 - 1 = 3
d) Since a = 4 is positive, the range of values of y = g(x) = 4 (x - 1)
2 - 1 is given by:
the inequality y ≥ k or y ≥ - 1
or the interval [-1, +∞)
The graph is shown below with the vertex at (1,-1) which is a minimum because a = 4 is positive graph opens upward.
Its x intercepts are located at (1/2,0) and (3/2,0) and its y intercept is located at (0,3).
The parabola has an axis of symmetry the vertical line x = 1.
The range of values of y is: y ≥ -1 as seen in the graph.
graph of a quadratic function in vertex form: g(x) = 4 (x - 1)<sup>2</sup> - 1

Figure 2. graph of quadratic function in vertex form: g(x) = 4 (x - 1)2 - 1

Example 3

Find the vertex of the graph of function h(x) = - (x + 1/2) 2 + 2, its axis of symmetry, the y intercept, the x intercepts and graph it. What is the range of value soy y = h(x)?

Solution to Example 3

Comparing h(x) = - (x + 1/2) 2 + 2 to h(x) = a(x - h) 2 + k, h = - 1/2, k = 2 and a = - 1.
The vertex is at the point (-1/2 , 2).
The axis of symmetry is the vertical line x = - 1/2
a = - /12 is negative , the parabola (graph of g) opens downward.
The y intercept is at (0 , h(0)) = (0 , 1.75)
The x intercept are found by solving
h(x) = - (x + 1/2)
2 + 2 = 0
The solutions to the above equation are: -1/2 - √2 and -1/2 + √2.
Two x intercepts are located at: (-1/2 - √2 , 0) and (-1/2 + √2 , 0)
The graph is shown below with the vertex at (-1/2 , 2) which is a maximum because a = -1/2 is negative and the graph opens downward.
The x intercepts are located at (-1/2 - √2 , 0)≈ (-1.91,0) and (-1/2 + √2 , 0) ≈ (0.91,0).
The y intercept is located at (0,1.75)
The axis of symmetry is the vertical line given by the equation: x = h = -1/2.
The leading coefficient a = - 1 is negative and k = 2, therefore the range of y = h(x) is given by: y ≤ 2 (check in the graph below).
graph of a quadratic function in vertex form: h(x) = - (x + 1/2)<sup>2</sup> + 2

Figure 3. graph of quadratic function in vertex form: h(x) = - (x + 1/2)2 + 2

Graph of quadratic functions in standard form: f(x) = a x2 + b x + c


One of the best method to graph quadratic functions given in standard form f(x) = a x
2 + b x + c is to first rewrite them in vertex form as follows
f(x) = a(x - h)
2 + k , vertex located at the point (h , k)
where h = - b / (2 a) and k = f(h).
then graph them using the vertex, x and y intercepts and some more points if needed.

Example 4

Find the vertex of the graph of function m(x) = x 2 + 2 x, its axis of symmetry, the y intercept, the x intercepts and graph it. What is the range of value soy y = m(x)?

Solution to Example 4

Identify the coefficients a, b and c of the given function
a = 1, b = 2 and c = 0
Use the formula for h and k
h = - b / (2 a) = - 2 / (2 1) = - 1
k = m(h) = (h)
2 + 2 (h) = (-1) 2 + 2 (-1) = - 1
Use h and k found above to write m(x) in vertex form as follows
m(x) = a (x - h)
2 + k = (x + 1) 2 - 1
Note: as an exercise, expand (x + 1)2 - 1 and see that it gives x2 + 2 x.
Hence vertex at (h , k) = (-1 , -1)
The axis of symmetry is the vertical line x = h = - 1
a = 1 is positive , the parabola (graph of m) opens upward.
The y intercept is at (0 , m(0)) = (0 , 0)
The x intercept are found by solving
m(x) = x
2 + 2 x = 0 (you may also solve the vertex form if it is easier)
The above equation is solved by factoring:
x
2 + 2 x = x(x + 2) = 0
2 solutions : x = 0 and x = - 2
Two x intercepts are located at: (0 , 0) and (- 2 , 0)
The leading coefficient a is positive and k = -1, therefore the range of y = m(x) is given by: y ≥ - 1 (check in the graph below).
The graph is shown below with the vertex at (- 1, -1) which is a minimum because a = 1 is positive graph opens upward, its x intercepts at (0 , 0) and (- 2 , 0). Its y intercept at (0 , 0) and axis of symmetry x = h = -1.
The range of y = m(x) is given by y ≥ -1.
graph of a quadratic function in standard form: m(x) = x<sup>2</sup> + 2 x

Figure 4. graph of quadratic function in standard form: m(x) = x2 + 2 x

Example 5

Find the vertex of the graph of function p(x) = - 2 x 2 + 3 x + 1, its axis of symmetry, the y intercept, the x intercepts and graph it. What is the range of value of y = p(x)?

Solution to Example 5

Identify the coefficients a, b and c of the given function
a = - 2, b = 3 and c = 1
Use the formula for h and k
h = - b / (2 a) = - 3 / (2 (-2)) = 3 / 4
k = p(h) = - 2 (h)
2 + 3 (h) + 1 = -2 (3/4) 2 + 3 (3/4) + 1 = 17/8 ≈ 2.13
p(x) is written in vertex form as follows
p(x) = a (x - h)
2 + k = - 2 (x - 3/4) 2 + 17/8
Note: as an exercise, expand - 2 (x - 3/4)2 + 17/8 and see that it gives - 2 x 2 + 3 x + 1.
Hence vertex at (h , k) = (3/4 , 17/8)
The axis of symmetry is the vertical line x = h = 3/4
a = - 2 is negative , the parabola (graph of p) opens downward.
The y intercept is at (0 , p(0)) = (0 , 1)
The x intercept are found by solving
p(x) = - 2 (x - 3/4)
2 + 17/8 = 0 (in this example it is easier to solve the vertex form)
The above equation is solved by extracting the square root:
2 (x - 3/4)
2 = 17/8
2 solutions : x = 3/4 + √(17/16) ≈ 1.78 and x = 3/4 - √(17/16) ≈ - 0.28
Two x intercepts are located at: (3/4 - √(17/16) , 0) ≈ (-0.28 , 0) and (3/4 + √(17/16) , 0) ≈ (1.78 , 0)
The graph is shown below with the vertex at (3/4, 17/8 ) which is a maximum because a = - 2 is negative and graph opens downward.
Its x intercepts at (-0.28 , 0) and (1.78 , 0). Its y intercept at (0 , 1) and axis of symmetry x = h = 3/4.
The leading coefficient a = - 2 is negative and k = 17/8, therefore the range of y = p(x) is given by: y ≤ 17/8 (≈2.13) (check in the graph below).
Two extra points are found to help graph p. (-1,p(-1)) = (-1,-4) and (3,p(3)) = (3 , -8)
graph of a quadratic function in standard form: p(x) = - 2 x <sup>2</sup> + 3 x + 1

Figure 5. graph of quadratic function in standard form: p(x) = - 2 x 2 + 3 x + 1

Exercises

Graph the quadratic functions given in standard form.
1) Find the vertex, x and y intercepts and graph s(x) = -(1/2)x
2 + x + 1. Find the range of y = s(x)
2) Find the vertex, x and y intercepts and graph t(x) = 4 x
2 + 4 x + 2. Find the range of y = t(x)

Solutions to Above Exercises

1) vertex at (1,3/2) , x intercepts at (1 + √3 , 0) and (1 - √3 , 0) , y intercept at (0 , 1) , range of y = s(x): y ≤ 3/2
graph of a quadratic function in standard form: s(x) = -(1/2)x<sup>2</sup> + x + 1

Figure 6. graph of quadratic function in standard form: s(x) = -(1/2)x2 + x + 1
2) vertex at (-1 , 0) , one x intercepts at (-1 , 0) which is also the vertex , y intercept at (0 , 2) , range of y = t(x): y ≥ 0
graph of a quadratic function in standard form: t(x) = 4 x<sup>2</sup> + 4 x + 2

Figure 7. graph of quadratic function in standard form: t(x) = 4 x2 + 4 x + 2

More References and links on quadratic functions and parabolas

Vertex and Intercepts Parabola Problems.
Find the Points of Intersection of a Parabola with a Line.
High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
More
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers Quadratus

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