Graph Quadratic Functions

Graphs of quadratic functions of the vertex form \( f(x) = a (x - h)^2 + k \) and of the standard form \( f(x) = a x^2 + b x + c \) 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.

Page Content

• Graph of the basic quadratic function: \( f(x) = x^2 \)
• Graph of quadratic functions in standard form: \( f(x) = a x^2 + b x + c \)
• Exercises
• Solutions to the Exercises
• References

Graph of the basic quadratic function: \( f(x) = x^2 \)

Example 1

We first note that

1) the square of real number is either positive or zero, \( x^2 \ge 0 \) 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 : \( y \ge 0 \)

Let us use a table to find points on the graph of function \( f \). \[ \begin{array}{|c|c|c|} \hline x & f(x) = x^2 \\ \hline 0 & 0^2 = 0 \\ \hline -1 & (-1)^2 = 1 \\ \hline 1 & 1^2 = 1 \\ \hline -2 & (-2)^2 = 4 \\ \hline 2& 2^2 = 4 \\ \hline -3& (-3)^2 = 9 \\ \hline 3 & 3^2 = 9 \\ \hline \end{array} \]

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 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 when \( |a| \gt 1 \) or compression \( |a| \lt 1 \) plus reflection on x axis if \( a \) is negative.

3) From the quadratic expression \( a(x - h)^2 \) to \( a(x - h)^2 + k \): a vertical shift by \( k \) units.

Properties of a quadratic functions of the vertex form

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 : \( y \ge k \) if \( a \) is positive or \( y \le k \) 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

Comparing \( g(x) = 4(x - 1)^2 - 1 \) to \( g(x) = a(x - h)^2 + k \), we can write: \( h = 1 \), \( k = -1 \) 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 \[ g(x) = 4(x - 1)^2 - 1 = 0 \] \[ 4(x - 1)^2 = 1 \] \[ 2(x - 1) = \pm \sqrt{1} \] \[ 2(x - 1) = 1 \text{ gives } x = \frac{3}{2} = 1.5 \] \[ 2(x - 1) = -1 \text{ gives } x = \frac{1}{2} = 0.5 \] 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 \( y \geq k \) or \( y \geq -1 \) or the interval \[ [-1, +\infty) \] 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: \( y \geq -1 \) as seen in the graph.

Example 3

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

Comparing the function \( h(x) = -\left(x + \frac{1}{2}\right)^2 + 2 \) to the standard form \( h(x) = a(x - h)^2 + k \), we identify: \( h = -\frac{1}{2} \), \( k = 2 \) and \( a = -1 \).

The vertex is at the point \( \left(-\frac{1}{2}, 2\right) \).

The axis of symmetry is the vertical line \( x = -\frac{1}{2} \).

Since \( a = -1 \) is negative, the parabola (graph of \( h \)) opens downward.

The y-intercept is at \( (0, h(0)) = (0, 1.75) \).

The x-intercepts are found by solving the equation:

Solving this, the solutions are:

So the x-intercepts are located at:

The graph below shows the vertex at \( \left(-\frac{1}{2}, 2\right) \), which is a maximum point because \( a = -1 \) is negative, so the parabola opens downward.

The approximate x-intercepts are:

The y-intercept is located at \( (0, 1.75) \).

The axis of symmetry is the vertical line given by the equation \( x = h = -\frac{1}{2} \).

Since the leading coefficient \( a = -1 \) is negative and \( k = 2 \), the range of \( y = h(x) \) is:

Refer to the graph below for a visual representation of the function's features.

Graph of quadratic functions in standard form: \( f(x) = a x^2 + b x + c \)

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 \( h = - \dfrac{b}{2 a}\) 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 formulas for \( h \) and \( k \):

\[ h = \frac{-b}{2a} = \frac{-2}{2 \times 1} = -1 \]

\[ k = m(h) = h^2 + 2h = (-1)^2 + 2(-1) = -1 \]

Use \( h \) and \( k \) found above to write \( m(x) \) in vertex form:

\[ m(x) = a(x - h)^2 + k = (x + 1)^2 - 1 \]

Note: As an exercise, expand \( (x + 1)^2 - 1 \) and verify that it simplifies to \( x^2 + 2x \).

Hence, the vertex is at \( (h, k) = (-1, -1) \).

The axis of symmetry is the vertical line \( x = h = -1 \).

Since \( a = 1 \) is positive, the parabola (graph of \( m \)) opens upward.

The y-intercept is at \( (0, m(0)) = (0, 0) \).

The x-intercepts are found by solving:

\[ m(x) = x^2 + 2x = 0 \]

This equation is solved by factoring:

\[ x^2 + 2x = x(x + 2) = 0 \]

Two solutions: \( x = 0 \) and \( x = -2 \).

So, the x-intercepts are located at \( (0, 0) \) and \( (-2, 0) \).

The leading coefficient \( a = 1 \) is positive and \( k = -1 \), so the range of \( y = m(x) \) is given by:

\[ y \geq -1 \]

The graph shows the vertex at \( (-1, -1) \), which is a minimum because \( a = 1 \) is positive, meaning the graph opens upward. The x-intercepts are at \( (0, 0) \) and \( (-2, 0) \), and the y-intercept is at \( (0, 0) \). The axis of symmetry is \( x = h = -1 \).

Therefore, the range of \( y = m(x) \) is \( y \geq -1 \).

Example 5

Solution to Example 5

Identifying the Coefficients of the Quadratic Function

The given quadratic function has the following coefficients: \( a = -2 \), \( b = 3 \) and \( c = 1 \).

We use the standard formulas for the vertex coordinates \( h \) and \( k \):

\[ h = \frac{-b}{2a} = \frac{-3}{2 \times (-2)} = \frac{3}{4} \]

\[ k = p(h) = -2(h)^2 + 3(h) + 1 = -2\left(\frac{3}{4}\right)^2 + 3\left(\frac{3}{4}\right) + 1 = \frac{17}{8} \approx 2.13 \]

The function \( p(x) \) can be written in vertex form:

\[ p(x) = a(x - h)^2 + k = -2\left(x - \frac{3}{4}\right)^2 + \frac{17}{8} \]

Note: As an exercise, try expanding \( -2\left(x - \frac{3}{4}\right)^2 + \frac{17}{8} \) to verify that it simplifies to \( -2x^2 + 3x + 1 \).

• Vertex: \( (h, k) = \left(\frac{3}{4}, \frac{17}{8}\right) \)
• Axis of symmetry: \( x = h = \frac{3}{4} \)
• Leading coefficient \( a = -2 \) is negative, so the parabola opens downward.
• Y-intercept: \( (0, p(0)) = (0, 1) \)

To find the x-intercepts, solve the equation \( p(x) = 0 \) using the vertex form:

\[ -2\left(x - \frac{3}{4}\right)^2 + \frac{17}{8} = 0 \]

Extracting the square root:

\[ 2\left(x - \frac{3}{4}\right)^2 = \frac{17}{8} \]

\[ \left(x - \frac{3}{4}\right)^2 = \frac{17}{16} \]

\[ x = \frac{3}{4} \pm \sqrt{\frac{17}{16}} \approx 0.75 \pm 1.03 \]

\[ x \approx -0.28 \quad \text{and} \quad x \approx 1.78 \]

The x-intercepts are approximately:

• \( (-0.28, 0) \)
• \( (1.78, 0) \)

Graph Characteristics

The graph of \( p(x) \) has the following key points:

• Vertex at \( \left(\frac{3}{4}, \frac{17}{8}\right) \) — this is a maximum since \( a = -2 \).
• X-intercepts at \( (-0.28, 0) \) and \( (1.78, 0) \)
• Y-intercept at \( (0, 1) \)
• Axis of symmetry at \( x = \frac{3}{4} \)

Since the leading coefficient \( a = -2 \) is negative and the vertex's y-value is \( \frac{17}{8} \), the range of the function is:

\[ y \leq \frac{17}{8} \approx 2.13 \]

Additional Points for Graphing

To assist with accurate graphing, two additional points on the parabola are:

• \( (-1, p(-1)) = (-1, -4) \)
• \( (3, p(3)) = (3, -8) \)

Exercises

Graph the quadratic functions given in standard form.

1) Find the vertex, x- and y-intercepts, and graph the function \( s(x) = -\frac{1}{2}x^2 + x + 1 \). Determine the range of the function \( y = s(x) \).

2) Find the vertex, x- and y-intercepts, and graph the function \( t(x) = 4x^2 + 4x + 2 \). Determine the range of the function \( y = t(x) \).

Solutions to Above Exercises

1)

• Vertex: \( (1, \frac{3}{2}) \)
• X-intercepts: \( \left(1 + \sqrt{3}, 0\right) \) and \( \left(1 - \sqrt{3}, 0\right) \)
• Y-intercept: \( (0, 1) \)
• Range: \[ y \leq \frac{3}{2} \]

2) The vertex of the parabola is at \( (-1, 0) \). There is one x-intercept at \( (-1, 0) \), which is also the vertex. The y-intercept occurs at \( (0, 2) \).

The range of the function \( y = t(x) \) is given by:

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