Quadratic functions take the shape of a parabola. Parabolas are utilised in bridges, radio telescopes, and car headlights, among other applications, due to their special shape properties. In fact, many famous landmarks are parabolas - for example, the Sydney Harbour Bridge.
Quadratic functions can be written in any of the following forms:
General Form: $y=ax^2+bx+c$y=ax2+bx+c
Factored or $x$x-intercept form: $y=a\left(x-\alpha\right)\left(x-\beta\right)$y=a(x−α)(x−β)
Turning point form: $y=a\left(x-h\right)^2+k$y=a(x−h)2+k
Each form has an advantage for different key features that can be identified quickly. However, for each form the role of $a$a remains the same.
That is:
The $x$x-intercepts are where the parabola crosses the $x$x-axis. This occurs when $y=0$y=0.
The $y$y-intercept is where the parabola crosses the $y$y-axis. This occurs when $x=0$x=0.
Parabolas can have none, one, or two $x$x-intercepts as shown below.
No solutions | One solution | Two solutions |
Maximum or minimum values are also known as the turning points, and they are found at the vertex of the parabola.
Parabolas that are concave up have a minimum value. This means the $y$y-value will never go under a certain value. |
Parabolas that are concave down have a maximum value. This means the $y$y-value will never go over a certain value. |
Maximum and minimum values occur on a parabola's axis of symmetry. This is the vertical line that evenly divides a parabola. We can use this symmetry to find the $x$x-value of the vertex by averaging the two $x$x-intercepts (if they exist). Alternatively, we can find the equation of the axis of symmetry and hence the $x$x coordinate of the vertex by using the general form of the quadratic $y=ax^2+bx+c$y=ax2+bx+c and substituting the relevant values of $a$a and $b$b into:
$x=\frac{-b}{2a}$x=−b2a
$y=ax^2+bx+c$y=ax2+bx+c
$y=a\left(x-\alpha\right)\left(x-\beta\right)$y=a(x−α)(x−β)
$y=a\left(x-h\right)^2+k$y=a(x−h)2+k
Consider the quadratic function $y=-x^2+4x+12$y=−x2+4x+12.
What is the concavity of the parabola?
Concave up
Concave down
What is the $y$y value of the $y$y-intercept?
Find the $x$x values of the $x$x-intercepts.
Find the equation of the axis of symmetry.
Find the coordinates of the vertex of the parabola.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Graph the function $y=-x^2+4x+12$y=−x2+4x+12.
Consider the parabola $y=\left(x+1\right)\left(x-3\right)$y=(x+1)(x−3).
Find the $y$y value of the $y$y-intercept.
Find the $x$x values of the $x$x-intercepts.
Write all solutions on the same line separated by a comma.
State the equation of the axis of symmetry.
Find the coordinates of the vertex.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Graph the parabola.
Consider the graph.
Determine the coordinates of the vertex of the parabola.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Given that the graph has equation of the form $y=a\left(x-h\right)^2+k$y=a(x−h)2+k, determine the equation of the parabola.
We may also be asked to find the equation of a parabola, given a sketch or relevant pieces of information. Using the vertex and factored forms of quadratic functions can be very helpful. Recall that every quadratic expression can be written in factored form as $a\left(x-\alpha\right)\left(x-\beta\right)$a(x−α)(x−β) for the $x$x-intercepts or $a\left(x-h\right)^2+k$a(x−h)2+k for the vertex, where $a$a is simply the coefficient from the general form $y=ax^2+bx+c$y=ax2+bx+c.
We can use the $x$x-intercepts or the coordinates of the vertex to put the equation in the relevant form (factored or vertex form), and then solve for $a$a by substituting the coordinates of any additional point.
Form a monic quadratic equation which has solutions $x=-3$x=−3 and $x=-5$x=−5. Express the equation in expanded form.
Consider the quadratic $y=x^2-12x+32$y=x2−12x+32
Express the equation in the form $y=a\left(x-h\right)^2+k$y=a(x−h)2+k by completing the square.
Find the zeros of the quadratic function.
Write both solutions on the same line, separated by a comma.
Find the coordinates of the vertex of the parabola represented by this quadratic function.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Use the zeros and the vertex found in the previous parts to plot the curve.