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Chapter Contents

• Quadratic Equations
• 1. Solving Quadratic Equations by Factoring
• 2. Completing the Square
• 3. The Quadratic Formula
• 4. The Graph of the Quadratic Function
• 5. Equations in Quadratic Form

• Comments, Questions?

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4. The Graph of the Quadratic Function

In general, the graph of a quadratic equation

y = ax² + bx + c

is a parabola.

[You can also see a more detailed description of parabolas in the Plane Analytic Geometry section.]

---

If a > 0, then the parabola has a minimum point and it opens upwards (U-shaped) eg.

y = x² + 2x − 3

---

If a < 0, then the parabola has a maximum point and it opens downwards (n-shaped) eg.

y = -2x² + 5x + 3

 

Sketching Parabolas

In order to sketch the graph of the quadratic equation, we follow these steps :

(a) Check if a > 0 or a < 0 to decide if it is U-shaped or n-shaped.

(b) The Vertex: The x-coordinate of the minimum point (or maximum point) is given by

(which can be shown using completing the square method, which we met earlier).

We substitute this x-value into our quadratic function (the y expression). Then we will have the (x, y) coordinates of the minimum (or maximum) point. This is called the vertex of the parabola.

(c) The coordinates of the y-intercept (substitute x = 0). This is always easy to find!

(d) The coordinates of the x-intercepts (substitute y = 0 and solve the quadratic equation), as long as they are easy to find.

Example 1

Sketch the graph of the function y = 2x² − 8x + 6

Let's see the graph of this quadratic function using LiveMath. Play and enjoy!

LIVEMath

Normal answer

We first identify that a = 2, b = -8 and c = 6.

Step (a) Since a = 2, a > 0 hence the function is a parabola with a minimum point and it opens upwards (U-shaped)

 

Step (b) The x co-ordinate of the minimum point is:

The y value of the minimum point is

y = 2(2)² - 8(2) + 6 = -2

So the minimum point is (2, -2)

 

Step (c) The y-intercept is found by substituting x = 0 into the y expression.

y = 2(0)² - 8(0) + 6 = 6

So (0, 6) is the y-intercept.

 

Step (d) The x-intercepts are found by setting y = 0 and solving:

2x² - 8x + 6 = 0

2(x² - 4x + 3) = 0

2(x - 1)(x - 3) = 0

So x = 1, or x = 3.

Using the above information, the sketch of the curve will be :

Before we go on, let's have a play with quadratic graphs using LiveMath.

 

Investigate Solutions of Quadratic Equations Using LiveMath

LIVEMath

Now see if you really know what's going on. Try this LiveMath challenge.

LIVEMath

Example 2

Sketch the graph of the function y = -x² + x + 6

Answer

 

Exercise

Sketch the graph y = -x² − 4x − 3

Answer