Previously, key features of linear equations have been explored including the gradient, the $x$x-intercept, and the $y$y-intercept. The graph of a quadratic equation is a parabola and is a curved or concave shape (either concave up or down, depending on the equation). The following sections describe the key features of a parabola.
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.
These intercepts are shown in the picture below:
There may be one or two or even no $x$x-intercepts:
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 into two sides down the middle. Using the general form of the quadratic $y=ax^2+bx+c$y=ax2+bx+c, substitute the relevant values of $a$a and $b$b into the following equation:
$x=\frac{-b}{2a}$x=−b2a
Does it look familiar? This is part of the quadratic formula. It is the half-way point between the two solutions.
Positive and negative gradients were previously explored for straight lines. If we draw the tangent line to a curve at a particular point then we can approximate the gradient to the curve. A parabola has a positive gradient in some places and negative gradient in others. The parabola's maximum or minimum value, at which the gradient is $0$0, is the division between the positive and negative gradient regions. Hence, why it is also called the turning point.
Look at the picture below. One side of the parabola has a positive gradient, there is a turning point with a zero gradient, and the other side of the parabola has a negative gradient.
In this particular graph, the gradient is positive when $x<1$x<1 and the parabola is increasing for these values of $x$x. Similarly, the gradient is negative when $x>1$x>1, and for these values of $x$x, the parabola is decreasing.
Examine the given graph and answer the following questions.
What are the $x$x values of the $x$x-intercepts of the graph? Write both answers on the same line separated by a comma.
What is the $y$y value of the $y$y-intercept of the graph?
What is the minimum value of the graph?
Examine the attached graph and answer the following questions.
What is the $x$x-value of the $x$x-intercept of the graph?
What is the $y$y value of the $y$y-intercept of the graph?
What is the absolute maximum of the graph?
Determine the interval of $x$x in which the graph is increasing.
Give your answer as an inequality.
As for linear equations, there are a number of forms for graphing quadratic functions:
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 following are examples of how to find all the key features for each form.
$y=a\left(x-h\right)^2+k$y=a(x−h)2+k
Consider the equation $y=-\left(x+2\right)^2+4$y=−(x+2)2+4.
Find the $x$x-intercepts. Write all solutions on the same line, separated by a comma.
Find the $y$y-intercept.
Determine the coordinates of the vertex.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Graph the equation.
Consider the equation $y=\left(x-3\right)^2+4$y=(x−3)2+4.
Does the graph have any $x$x-intercepts?
No
Yes
Find the $y$y-intercept.
Determine the coordinates of the vertex.
Vertex $=$=$\left(\editable{},\editable{}\right)$(,)
Graph the equation.
$y=a\left(x-\alpha\right)\left(x-\beta\right)$y=a(x−α)(x−β)
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.
$y=ax^2+bx+c$y=ax2+bx+c
Alternatively, use the method of completing the square to rewrite the quadratic in turning point form.
Sometimes, the solutions for a quadratic may be asked for. This is referring to the $x$x-intercepts which satisfy the equation $y=0$y=0. The solutions are sometimes called the roots of the equation.
Consider the quadratic function $y=x^2+2x-8$y=x2+2x−8.
Determine the $x$x-value(s) of the $x$x-intercept(s) of this parabola. Write all answers on the same line separated by commas.
Determine the $y$y-value of the $y$y-intercept for this parabola.
Determine the equation of the vertical axis of symmetry for this parabola.
Find the $y$y-coordinate of the vertex of the parabola.
Draw a graph of the parabola $y=x^2+2x-8$y=x2+2x−8.
A parabola has the equation $y=x^2+4x-1$y=x2+4x−1.
Express the equation of the parabola in the form $y=\left(x-h\right)^2+k$y=(x−h)2+k by completing the square.
Find the $y$y-intercept of the curve.
Find the vertex of the parabola.
Vertex $=$= $\left(\editable{},\editable{}\right)$(,)
Is the parabola concave up or down?
Concave up
Concave down
Hence plot the curve $y=x^2+4x-1$y=x2+4x−1
Consider the graph of the function $f\left(x\right)=-x^2-x+6$f(x)=−x2−x+6.
Using the graph, write down the solutions to the equation $-x^2-x+6=0$−x2−x+6=0.
If there is more than one solution, write the solutions separated by commas.
Technology can be used to graph and find key features of a quadratic function. We may need to use our knowledge of the graph or calculate the location of some key features to find an appropriate view window.
Use your calculator or other handheld technology to graph $y=4x^2-64x+263$y=4x2−64x+263.
Then answer the following questions.
What is the vertex of the graph?
Give your answer in coordinate form.
The vertex is $\left(\editable{},\editable{}\right)$(,)
What is the $y$y-intercept?
Give your answer in coordinate form.
The $y$y-intercept is $\left(\editable{},\editable{}\right)$(,)