Online Mathematics Book

Quadratic functions are all of the form: $f(x) = ax^2+bx+c$ where $$a$$, $$b$$ and $$c$$ are known as the quadratic's coefficients and are all real numbers, with $$a\neq 0$$.

The Parabola

Given a quadratic function $$f(x) = ax^2+bx+c$$, it is described by its curve: $y = ax^2+bx+c$ This type of curve is known as a parabola. A typical parabola is shown here:

Concave-Up & Concave-Down: the Role of $$a$$

Given a parabola $$y=ax^2+bx+c$$, depending on the sign of $$a$$, the $$x^2$$ coefficient, it will either be concave-up or concave-down:

• $$a>0$$: the parabola will be concave-up
• $$a<0$$: the parabola will be concave-down

We illustrate each of these two cases here:

The parabola $$y=2x^2 - 12x+9$$.
The $$x^2$$ coefficient is $$2$$, which is positive.
This corresponds to the $$a>0$$ scenario stated above.
The parabola is shown here:

The parabola $$y=-x^2 +4x + 4$$.
The $$x^2$$ coefficient is $$-1$$, which is negative.
This corresponds to the $$a< 0$$ scenario stated above.
The parabola is shown here:

$$y$$-intercept, $$c$$

Given a parabola $$y=ax^2+bx+c$$, the point at which it cuts the $$y$$-axis is known as the $$y$$-intercept.

The $$y$$-intercept will always have coordinates: $\begin{pmatrix}0,c\end{pmatrix}$ where $$c$$ is the only term in the parabola's equation without an $$x$$.

Example

The parabola defined by: $y = x^2+2x-3$ has $$y$$-intercept at: $\begin{pmatrix}0,-3\end{pmatrix}$ Where $$-3$$ is the only term without an $$x$$ in the parabola's equation.

This can be seen on this parabola's graph:

This can be confirmed algebraically we can find the $$y$$-intercept using the fact that when the curve cuts the $$y$$-axis: $$x=0$$, so replacing $$x$$ by $$0$$ in the parabola's equation leads to: $y = 0^2+2\times 0-3 = -3$ So the $$y$$-intercept is $$\begin{pmatrix}0,-3 \end{pmatrix}$$.

Axis of Symmetry

All parabola have a vertical axis of symmetry, with equation: $x = \frac{-b}{2a}$

Tutorial: Axis of Symmetry

In the following tutorial we see how to use the formula for finding a parabola's axis of symmetry.

Example

The parabola defined by: $y = x^2 - 6x+5$ has axis of symmetry: \begin{aligned} x & = \frac{-(-6)}{2\times 1} \\ & = \frac{6}{2} \\ x & = 3 \end{aligned} This parabola's vertical axis of symmetry has equation: $x = 3$ This is illustrated in the graph we see here, where the axis of symmetry is the dotted line.

Vertex of a Parabola

Given a quadratic function $$f(x) = ax^2+bx+c$$, depending on the sign of the $$x^2$$ coefficient, $$a$$, its parabola has either a minimum or a maximum point:

• if $$a>0$$: it has a maximum point
• if $$a<0$$: it has a minimum point
in either case the point (maximum, or minimum) is known as a vertex.

Finding the Vertex

To find the vertex we calculate its $$x$$-coordinate, $$h$$, with the formula given below.
We then calculate its $$y$$-coordinate, $$k$$, by plugging in the

• Step 1: calculate the $$x$$-coordinate of the vertex, $$h$$, using the formula: $h = \frac{-b}{2a}$
• Step 2: calculate the $$y$$-coordinate of the vertex, $$k$$, by replacing $$x$$ inside $$y=ax^2+bx+c$$ and calculating the value of $$y$$.

Tutorial: Coordinates of the Vertex

In the following tutorial we learn how to find the coordinates of a parabola's vertex, in other words the coordinates of its maximum, or minimum, point.

Example

Consider quadratic function whose parabola is described by: $y = 2x^2 - 4x - 6$

1. State whether this parabola's vertex is a maximum, or a minimum.
2. Find the coordinates of this parabola's vertex.

Solution

1. The $$x^2$$ coefficient is $$2$$. Since $$2>0$$ this parabola's vertex is a minimum point.
2. To find the coordinates of the vertex, we follow the two steps we read further-up:
• Step 1: we calculate the $$x$$-coordinate, $$h$$, of the vertex using the formula: $h = \frac{-b}{2a}$ Looking at $$y=2x^2 - 4x-6$$, we see that: $a = 2, \ b = -4, \ c = -6$ So the formula for the $$x$$-coordinate becomes: \begin{aligned} h & = \frac{-b}{2a}\\ & = \frac{-(-4)}{2\times 2} \\ & = \frac{4}{4} \\ h & = 1 \end{aligned} So the $$x$$-coordinate of the vertex is $$h=1$$.
• Step 2: we calculate the $$y$$-coordinate of the vertex by replacing $$x$$ by $$1$$ inside $$y=2x^2-4x-6$$ and calculating the value of $$y$$.

That's: \begin{aligned} y & = 2\times 1^2-4\times 1-6 \\ & = 2\times 1 - 4 - 6 \\ & = 2 - 4 - 6 \\ & = -2 - 6 \\ y & = -8 \end{aligned} So the $$y$$-coordinate of the vertex is $$k=-8$$.
Finally we can state that this parabola has a minimum point with coordinates $$\begin{pmatrix}1, -8 \end{pmatrix}$$.
This result is confirmed when we look at this parabola's graph. We can clearly see its vertex is a minimum point, with coordinates $$\begin{pmatrix}1, -8 \end{pmatrix}$$:

Domain & Range

Domain

Given a quadratic function, $$f(x)=ax^2+bx+c$$, and its parabola, $$y=ax^2+bx+c$$, unless otherwise stated, the domain is:

All Real Numbers, $$\mathbb{R}$$.
We can write this as: $\text{Domain} = \mathbb{R}$

Range

The range depends on two things:

• the sign of the coefficient $$a$$.
• the $$y$$-coordinate, $$k$$, of the vertex.
• if $$a>0$$: the parabola's vertex is a minimum point and the range will be: $\text{Range}: y\geq k$
• if $$a< 0$$: the parabola's vertex is a maximum point and the range will be: $\text{Range}: y\leq k$

Example

Find the range of each of the quadratic functions defined by:

1. The parabola $$y = 2x^2 - 8x+11$$
2. The parabola $$y = -x^2 - 6x - 5$$

Solution

1. To find the range, we find the coordinates of the vertex of $$y = 2x^2 - 8x+11$$ (either using a graphical calculator, or algebraically).
We find that the parabola has a minimum point with coordinates $$\begin{pmatrix}2,3\end{pmatrix}$$.
This can be seen on the parabola shown here:
Since the parabola is concave-up, the range is: $\text{Range}: \ y \geq 3$

2. To find the range, we find the coordinates of the vertex of $$y = -x^2 - 6x - 5$$ (either using a graphical calculator, or algebraically).
We find that the parabola has a maximum point with coordinates $$\begin{pmatrix}-3,4\end{pmatrix}$$.
This can be seen on the parabola shown here:
Since the parabola is concave-down, the range is: $\text{Range}: \ y \leq 4$

$$x$$-intercepts

Given a quadratic function $$f(x) = ax^2+bx+c$$, its parabola $$y=ax^2+bx+c$$ cuts the $$x$$-axis either:

• Twice
• Once
• Not at all
The $$x$$-values at which the curve cuts the $$x$$-axis are found by solving the quadratic equation: $ax^2+bx+c = 0$ If you're unsure of how to solve this type of equation, make sure to read through our notes on the quadratic formula.

Example

Find the $$x$$-intercept(s) for each of the following parabola:

1. $$y = 3x^2 - 3x-6$$
2. $$y = -x^2-8x-16$$
3. $$y = 2x^2-4x+5$$

Solution

1. The parabola $$y = 3x^2 - 3x-6$$ cuts the $$x$$-axis when $3x^2-3x-6 = 0$ We solve this equation in two steps
• Step 1: calculate the discriminant $$\Delta$$: \begin{aligned} \Delta & = b^2 - 4ac\\ & = (-3)^2-4\times 3\times (-6)\\ & = 9 - 12 \times (-6) \\ & = 9 - (-72)\\ & = 9 + 72 \\ \Delta & = 81 \end{aligned}
• Step 2: we now solve the equation, according to the sign of $$\Delta$$.

Since $$\Delta >0$$ the equation has two solutions, given by the formula: $x = \frac{-b\pm \sqrt{\Delta}}{2a}$ replacing $$\Delta$$, $$a$$ and $$b$$ by their respective values leads to: \begin{aligned}x &= \frac{-b\pm \sqrt{\Delta}}{2a} \\ & = \frac{-(-3) \pm \sqrt{81}}{2\times 3} \\ & = \frac{3 \pm 9}{6} \\ x & = \frac{1 \pm 3}{2} \\ \end{aligned} So the equation $$3x^2 - 3x-6 = 0$$ has two solutions: \begin{aligned} x & = \frac{1 - 3}{2} \\ & = \frac{-2}{2} \\ x & = -1 \end{aligned}
and
\begin{aligned} x & = \frac{1 + 3}{2} \\ & = \frac{4}{2} \\ x & = 2 \end{aligned} The solutions are $$x = -1$$ and $$x = 2$$.
This means that the parabola cuts the $$x$$-axis twice, at: $x = \begin{pmatrix}-1,0\end{pmatrix}$
and
$x = \begin{pmatrix}2,0\end{pmatrix}$ These two points are known as the parabola's $$x$$-intercepts.
Looking at the parabola $$y = 3x^2 - 3x-6$$ on an $$xy$$-grid we can confirm these results:
2. The parabola $$y = -x^2-8x-16$$ cuts the $$x$$-axis when: $-x^2-8x-16 = 0$ We solve this equation in two-steps:
• Step 1: calculate the discriminant $$\Delta$$: \begin{aligned} \Delta & = b^2 - 4ac\\ & = (-8)^2-4\times (-1)\times (-16\) \\ & = 64 - (-4)\times (-16)\\ & = 64 - 64 \\ \Delta & = 0 \end{aligned}
• Step 2: we now solve the equation, according to the sign of $$\Delta$$.

Since $$\Delta = 0$$ the equation has one solution, given by the formula: $x = \frac{-b}{2a}$ replacing $$a$$ and $$b$$ by their respective values leads to: \begin{aligned} x & = \frac{-b}{2a} \\ & = \frac{-(-8)}{2\times (-1)} \\ & = \frac{8}{-2} \\ x & = -4 \end{aligned} The solution to this equation is $$x = -4$$.
This means that the parabola cuts the $$x$$-axis once, at: $x = \begin{pmatrix}-4,0\end{pmatrix}$ This point is the parabola's $$x$$-intercept.
Looking at the parabola $$y = -x^2-8x-16$$ on an $$xy$$-grid we can confirm this result:
3. The parabola $$y = 2x^2-4x+5$$ cuts the $$x$$-axis when: $2x^2-4x+5 = 0$ We solve this equation in two-steps:
• Step 1: calculate the discriminant $$\Delta$$: \begin{aligned} \Delta & = b^2 - 4ac\\ & = (-4)^2 - 4\times 2\times 5 \\ & = 16 - 8 \times 5 \\ & = 16 - 40 \\ \Delta & = -24 \end{aligned}
• Step 2: we now solve the equation, according to the sign of $$\Delta$$.

Since $$\Delta < 0$$ the equation has no real solution
This means that the parabola does not cut/touch the $$x$$-axis.
Looking at the parabola $$y = 2x^2-4x+5$$ on an $$xy$$-grid we can confirm this result: