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 \(a\neq 0\).
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:
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:
We illustrate each of these two cases here:
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 \(x\).
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 \(\begin{pmatrix}0,-3 \end{pmatrix}\).
All parabola have a vertical axis of symmetry, with equation: \[x = \frac{-b}{2a}\]
In the following tutorial we see how to use the formula for finding a parabola's axis of symmetry.
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.
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:
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
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.
Consider quadratic function whose parabola is described by: \[y = 2x^2 - 4x - 6\]
Given a quadratic function, \(f(x)=ax^2+bx+c\), and its parabola, \(y=ax^2+bx+c\), unless otherwise stated, the domain is:
The range depends on two things:
Find the range of each of the quadratic functions defined by:
Given a quadratic function \(f(x) = ax^2+bx+c\), its parabola \(y=ax^2+bx+c\) cuts the \(x\)-axis either:
Find the \(x\)-intercept(s) for each of the following parabola: