# Curve Sketching 2: Graphing Quadratic Functions

In Curve Sketching 1, we have learned four easy ways to graph linear functions. In this post, we will learn how to sketch the parabola, the graph of quadratic functions. Quadratic Functions are functions with equation $y = ax^2 + bx +c$ where $a \neq 0$. This is the standard form. This equation can also be expressed in the form $y = a(x-h)^2 + k$ where $a \neq 0$ is the vertex form.

Unlike linear functions, we need more than two points to sketch the graph of quadratic functions. In the following discussion, we will examine the different properties of quadratic function and use them to sketch its graph.

Here are some of the properties of $y = ax^2 + bx + c$ in relation to its graph. For students, it is really important that you think about them — don’t just memorize.

1. Direction of the Graph

The graph of a quadratic function can either be $\cup$-shaped or $\cap$-shaped.  If $a >0$ then the graph is $\cup$-shaped, and if $a < 0$, the graph is $\cap$-shaped.

2. Vertex

In a quadratic function, its vertex is its minimum (lowest point) if the graph is $\cup$-shaped and its maximum (highest point) if the graph is $\cap$shaped.  This will give us a rough idea of how the graph will look like if we know the vertex and another point on the graph. In a function with equation in vertex form, the vertex of the parabola is at $(h,k)$.

3. Intercepts

The x and y intercepts of Quadratic function are also clues on how the graph looks like. The quadratic function can have at most two x intercepts (see The Geometric Meaning of Discriminant) and exactly one y-intercept.

4. Axis of Symmetry

The axis of symmetry of the quadratic function is a vertical line that passes through its vertex. In standard form is $\frac{-b}{2a}$, while in vertex form is $x = h$.

Worked Example

Using the properties above, let us sketch the graph of $y = 3x^2 - 6x + 1$.

Direction of Graph

The graph is $\cup$-shaped since $a = 3$ is greater than $0$.

Vertex

To get the vertex, we convert the equation into standard form which gives us $y = 3(x-1)^2 - 2$. This means that the vertex $(h,k) = (1,-2)$.

Intercepts

For the y-intercept, we set $x = 0$, which gives us $y = 1$. So, the graph passes through $(0,1)$.

The x-intercepts or the roots can be obtained by setting $y = 0$ giving us $3x^2 - 6x - 1 = 0$. Using the quadratic formula, we get $x \approx 1.82$ and $x \approx 0.18$.  So, the points also passes through $(1.82,0)$ and $(0.18,0)$. We have enough points to graph as shown above.

Axis of Symmetry

Since the parabola is symmetric, the axis of symmetry can be used to reflect the point on the y-axis resulting to the fifth point on the graph.

With all the points, it is now possible to sketch the graph as shown below.

At times, the quadratic function can have no x-intercepts making only the vertex and y-intercept as critical points. Of course, additional points on the graph may be obtained by substituting values for x.