2.07 Quadratic graphs

Graphing quadratics

As for linear equations, there are a number of forms for graphing quadratic functions:

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:

• If $a<0$a<0 then the quadratic is concave down
• If $a>0$a>0 then the quadratic is concave up
• The larger the magnitude of $a$a, the narrower the curve

 

The following are examples of how to find all the key features for each form.

Graphing from turning point form

$y=a\left(x-h\right)^2+k$y=a(x−h)2+k

• This form is given this name as the turning point can be read directly from the equation: $(h,k)$(h,k)
• Obtain this graph from the graph of $y=x^2$y=x2, by dilating (stretching) the graph by a factor of $a$a from the $x$x-axis, then translating the graph $h$h units horizontally and $k$k units vertically
• The axis of symmetry will be at $x=h$x=h
• The $y$y-intercept can be found by substituting $x=0$x=0 into the equation
• The $x$x-intercept can be found by substituting $y=0$y=0 into the equation and rearranging

 

Practice questions

Question 1

Question 2

 

Graphing from factored form

$y=a\left(x-\alpha\right)\left(x-\beta\right)$y=a(x−α)(x−β)

• From this form, read the $x$x-intercepts directly: $(\alpha,0)$(α,0) and $(\beta,0)$(β,0). If  $y=0$y=0, solve the equation using the null factor law.
• The axis of symmetry will be at the $x$x-coordinate midway between the two $x$x-intercepts. Alternatively, expand this equation to general form and then use the formula for the axis of symmetry: $x=\frac{-b}{2a}$x=−b2a​.
• The axis of symmetry is also the $x$x-coordinate of the turning point. Substitute this value into the equation then find the $y$y-coordinate of the turning point.
• The $y$y-intercept can be found by substituting $x=0$x=0 into the equation

 

Practice question

Question 3

 

Graphing from general form

$y=ax^2+bx+c$y=ax2+bx+c

• From this form, read the $y$y-intercept directly: $(0,c)$(0,c) 
• Find the axis of symmetry using the formula: $x=\frac{-b}{2a}$x=−b2a​
• The axis of symmetry is also the $x$x-coordinate of the turning point. Substitute this value into the equation, then find the $y$y-coordinate of the turning point.
• The $x$x-intercept can be found by substituting $y=0$y=0 into the equation and then using one of the methods for solving quadratics, such as factorising or the quadratic formula, to find the zeros (x - intercepts) of the equation.

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.

 

Practice questions

Question 4

question 5

Question 6

 

Graphing quadratics using technology

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.

Practice question

Question 7

 

Finding roots using the bisection method

The word bisect means to divide into two parts. There exists a method for estimating the solutions (or roots) of a polynomial equation that uses the idea of dividing an interval into parts. This is the bisection method.

For example, imagine the quadratic $f(x)=3x^2-4x-2$f(x)=3x2−4x−2. The graph of this function cuts the x-axis somewhere between $x=1$x=1 and $x=2$x=2, as shown:

Divide this interval in half and substitute this $x$x-value into the function. In this example, substitute $x=1.5$x=1.5 into $f(x)$f(x). If $f(1.5)=0$f(1.5)=0, then this is the root! If not, check the sign of the answer and compare to the sign of the function at either end of the interval.

From the graph, it is clear that $f(1.5)<0$f(1.5)<0 while $f(2)>0$f(2)>0. This means that the root, where $f(a)=0$f(a)=0, must be somewhere in between. This becomes the new search area for the bisection method algorithms to repeat these steps:

Now, repeat the steps using the endpoints of $x=1.5$x=1.5 and $x=2$x=2. Testing the midpoint of this interval, $f(1.75)>0$f(1.75)>0. Since $f(2)>0$f(2)>0 also, then the new interval must be between $x=1.5$x=1.5 and $x=1.75$x=1.75:

Keep repeating this process until the desired accuracy of the estimate. This could also be done using a table of values, or technology. 

For example, for the function $f(x)=3x^2-4x-2$f(x)=3x2−4x−2, applying the bisection method for endpoints $x=a$x=a and $x=b$x=b, and midpoint $x=c$x=c, at any stage gives the following table:

Step (n)	a	b	f(a)	f(b)	c=(a+b)/2	f(c)	Update a or b
0	1	2	-3	2	1.5	-1.25	a = c
1	1.5	2	-1.25	2	1.75	0.1875	b = c
2	1.5	1.75	-1.25	0.1875	1.625	-0.57813	a = c
3	1.625	1.75	-0.578125	0.1875	1.6875	-0.20703	a = c
4	1.6875	1.75	-0.20703125	0.1875	1.71875	-0.0127	a = c
5	1.71875	1.75	-0.0127	0.1875	1.734375	0.08667	b = c
6	1.71875	1.734375	-0.0127	0.08667	1.7265625	0.036804	b = c
7	1.71875	1.7265625	-0.0127	0.036804	1.722656	0.012009	b = c

Notice that a choice is made at each stage whether to replace $a$a or $b$b with the value of $c$c depending on the sign of $f(c)$f(c).

Here, $f(c)=-0.012$f(c)=−0.012 in the last line of the table. If a specific tolerance is chosen for $f(c)$f(c) to be "close enough" to $f(c)=0$f(c)=0, then this value of $x$x is chosen as the estimate for the root. A number of iterations might also be chosen. Here, the process has gone through 7 iterations. So, in this example, if $f(c)$f(c) was considered "close enough" to $f(c)=0$f(c)=0, then the estimate for the root would be $x=1.723$x=1.723 (correct to 3 decimal places).

Since this method is recursive (has repeated and predictable steps), it is well automated with technology like spreadsheets.

Experiment with different polynomial functions in the applet below to see the binomial method in action. Click and drag the intervals around to see the changes in the estimates for the roots.

 

Practice questions

QUESTION 8