Chapters

What Is a Quadratic Function?
How Do You Solve and Graph a Quadratic Function?
Proposed Exercises

Before jumping into the exercises, let's review the basics.

The best Mathematics tutors available

What Is a Quadratic Function?

A quadratic function is a second-degree polynomial function with the form: $\text{[math]}$ , where $\text{[math]}$ , $\text{[math]}$ , and $\text{[math]}$ are real constants, and $\text{[math]}$ .

The graph of a quadratic function is one of the conic sections (circle, ellipse, parabola, or hyperbola), but in this section, we’ll focus only on parabolas.

The graph of $\text{[math]}$ — the simplest quadratic function — reveals several key features of a parabola. For instance, $\text{[math]}$ , and $\text{[math]}$ for any other real value of $\text{[math]}$ . This means the function has a minimum at the point $\text{[math]}$ , which is called the vertex of the parabola.

If $\text{[math]}$ , the parabola opens upward (the vertex is at the bottom).
If $\text{[math]}$ , the parabola opens downward (the vertex is at the top).

How Do You Solve and Graph a Quadratic Function?

There are two main methods to solve and graph a quadratic function. Below are the steps for each:

✅ Vertex Formula Method

1 - Identify the values of $\text{[math]}$ , $\text{[math]}$ , and $\text{[math]}$ .
2 - Find the $\text{[math]}$ -value of the vertex using the vertex formula.
3 - Find the $\text{[math]}$ -value by plugging $\text{[math]}$ into the function.
4 - Write the vertex coordinates as $\text{[math]}$ .

✅ Completing the Square Method

1 - Write out the equation.
2 - Divide all terms by the coefficient of $\text{[math]}$ .
3 - Move the constant term to the right-hand side.
4 - Complete the square on the left-hand side.
5 - Factor the left-hand side.
6 - Solve and write the vertex coordinates $\text{[math]}$ .

Proposed Exercises

Score:

Solve and Graph the Following Quadratic Functions

$\text{[math]}$

Solution

1 Vertex
We apply the vertex formula:

$\text{[math]}$

So, the vertex is $\text{[math]}$ .

2 X-axis intercepts
We set the function equal to zero and calculate the solutions:

$\text{[math]}$

Which has no real solutions.

So, there are no intersections with the $\text{[math]}$ axis.

3 Y-axis intercept
$\text{[math]}$

So, the intersection with the $\text{[math]}$ axis is $\text{[math]}$ .

4 With the above information, the graphical representation is:

Parabola vertical

$\text{[math]}$

Solution

1 Vertex
We apply the vertex formula:

$\text{[math]}$

So, the vertex is $\text{[math]}$ .

2 X-axis intercepts
We set the function equal to zero and calculate the solutions:

$\text{[math]}$

So, the intersections with the $\text{[math]}$ axis are $\text{[math]}$ .

3 Y-axis intercept
$\text{[math]}$

So, the intersection with the $\text{[math]}$ axis is $\text{[math]}$ .

4 With the above information, the graphical representation is:

Parabola con vertice en el eje x

$\text{[math]}$

Solution

1 Vertex
We apply the vertex formula:

$\text{[math]}$

So, the vertex is $\text{[math]}$ .

2 X-axis intercepts
We set the function equal to zero and calculate the solutions:

$\text{[math]}$

We get the solutions $\text{[math]}$

So, the intersections with the $\text{[math]}$ axis are $\text{[math]}$ and $\text{[math]}$ .

3 Y-axis intercept
$\text{[math]}$

So, the intersection with the $\text{[math]}$ axis is $\text{[math]}$ .

4 With the above information, the graphical representation is:

Gráfica de una parábola hacia abajo

$\text{[math]}$

Solution

1 Vertex
We apply the vertex formula:

$\text{[math]}$

So, the vertex is $\text{[math]}$ .

2 X-axis intercepts
We set the function equal to zero and calculate the solutions:

$\text{[math]}$

We get the solution $\text{[math]}$

So, the intersection with the $\text{[math]}$ axis is $\text{[math]}$ .

3 Y-axis intercept
$\text{[math]}$

So, the intersection with the $\text{[math]}$ axis is $\text{[math]}$ .

4 With the above information, the graphical representation is:

Gráfica de una parábola hacia arriba

$\text{[math]}$

Solution

1 Vertex
We apply the vertex formula:

$\text{[math]}$

So, the vertex is $\text{[math]}$ .

2 X-axis intercepts
We set the function equal to zero and calculate the solutions:

$\text{[math]}$

Since the discriminant is negative, $\text{[math]}$ , there are no intersections with the $\text{[math]}$ axis.

3 Y-axis intercept
$\text{[math]}$

So, the intersection with the $\text{[math]}$ axis is $\text{[math]}$ .

4 With the above information, the graphical representation is:

Gráfica de una parábola que habré hacia arriba

Find the vertex and the equation of the axis of symmetry for the following parabolas

1 $\text{[math]}$ ;

2 $\text{[math]}$ ;

3 $\text{[math]}$ ;

4 $\text{[math]}$ ;

5 $\text{[math]}$ ;

6 $\text{[math]}$ ;

7 $\text{[math]}$ ;

8 $\text{[math]}$ ;

9 $\text{[math]}$ ;

10 $\text{[math]}$

Solution

The vertex of the parabola $\text{[math]}$ is given by $\text{[math]}$ , and the axis of symmetry is $\text{[math]}$ . For the parabola $\text{[math]}$ , the vertex is given by $\text{[math]}$

1 $\text{[math]}$

$\text{[math]}$

2 $\text{[math]}$

$\text{[math]}$

3 $\text{[math]}$

$\text{[math]}$

4 $\text{[math]}$

$\text{[math]}$

5 $\text{[math]}$

$\text{[math]}$

6 $\text{[math]}$

$\text{[math]}$

7 $\text{[math]}$

$\text{[math]}$

8 $\text{[math]}$

$\text{[math]}$

9 $\text{[math]}$

$\text{[math]}$

10 $\text{[math]}$

$\text{[math]}$

Without graphing, indicate how many times the following parabolas intersect the x-axis

1 $\text{[math]}$ ;

2 $\text{[math]}$ ;

3 $\text{[math]}$ ;

4 $\text{[math]}$ ;

5 $\text{[math]}$ .

Solution

We apply the discriminant $\text{[math]}$ , and based on its sign, determine whether the parabolas intersect the x-axis twice, once, or not at all.

1. $\text{[math]}$

We calculate the discriminant:

$\text{[math]}$

We calculate the discriminant:

$\text{[math]}$

Since the discriminant is positive, there are two points of intersection.

3. $\text{[math]}$

We calculate the discriminant:

$\text{[math]}$

We calculate the discriminant:

$\text{[math]}$

Since the discriminant is zero, there is one point of intersection.

5. $\text{[math]}$

We calculate the discriminant:

$\text{[math]}$

Since the discriminant is positive, there are two points of intersection.

A quadratic function has the form $\text{[math]}$ and passes through the point $\text{[math]}$ . Find the value of $\text{[math]}$ .

Solution

1 We substitute the point into the function

$\text{[math]}$

A quadratic function has the form $\text{[math]}$ and passes through the point $\text{[math]}$ . Find the value of $\text{[math]}$ .

Solution

1 We substitute the point into the function:
$\text{[math]}$

2 We solve for $\text{[math]}$ :
$\text{[math]}$

A quadratic function has the form $\text{[math]}$ and passes through the points $\text{[math]}$ , and $\text{[math]}$ . Find $\text{[math]}$ , and $\text{[math]}$ .

Solution

1 We substitute each point into $\text{[math]}$ :
$\text{[math]}$

2 We obtain the following system of equations:
$\text{[math]}$

3 Solving the system, we get $\text{[math]}$

A parabola has its vertex at the point $\text{[math]}$ and passes through the point $\text{[math]}$ . Find its equation.

Solution

1 The equation is written in the form $\text{[math]}$
2 We substitute the values of the vertex:
$\text{[math]}$

3 We substitute the point $\text{[math]}$ and solve for $\text{[math]}$ :
$\text{[math]}$

4 We substitute the value of $\text{[math]}$ :
$\text{[math]}$

A parabola has its vertex at the point $\text{[math]}$ and passes through the point $\text{[math]}$ . Find its equation.

Solution

1 The equation is written in the form $\text{[math]}$
2 We substitute the values of the vertex:
$\text{[math]}$

3 We substitute the point $\text{[math]}$ and solve for $\text{[math]}$ :
$\text{[math]}$

4 We substitute the value of $\text{[math]}$ and expand:
$\text{[math]}$

Starting from the graph of the function $\text{[math]}$ , represent:

1 $\text{[math]}$ ;

2 $\text{[math]}$ ;

3 $\text{[math]}$ ;

4 $\text{[math]}$ ;

5 $\text{[math]}$ ;

6 $\text{[math]}$ ;

7 $\text{[math]}$ ;

8 $\text{[math]}$ ;

9 $\text{[math]}$ ;

10 $\text{[math]}$ ;

Solution

We will use the graph of $\text{[math]}$ .

Parábola hacia arriba desde el origen

1 $\text{[math]}$

We reflect it over the x-axis and shift the graph of $\text{[math]}$ so the vertex is at $\text{[math]}$

Parabola negativa

2 $\text{[math]}$

We reflect it over the x-axis and shift the graph of $\text{[math]}$ so the vertex is at $\text{[math]}$

Parabola negativa 2

3 $\text{[math]}$

We reflect it over the x-axis and shift the graph of $\text{[math]}$ so the vertex is at $\text{[math]}$

Parabola trasladada horizontalmente 1

4 $\text{[math]}$

We reflect it over the x-axis and shift the graph of $\text{[math]}$ so the vertex is at $\text{[math]}$

Parabola trasladada horizontalmente 2