It can be helpful to sketch a graph of a polynomial function in order to understand its behaviour over different parts of its domain. To do this, we gather information about the function at certain key points and then infer the shape of the graph from that information.
Given a polynomial function $P$P defined on a subset of the real numbers, we can easily find (by substitution) the value $P(0)$P(0) if it exists. This will be the intercept of the graph on the vertical axis.
The graph cuts the horizontal axis at the zeros of the function. These are the roots of the equation $P(x)=0$P(x)=0. A polynomial of odd degree in a real number $x$x always has at least one zero but a polynomial function of even degree may have no zeros. In either case, the zeros can be found algebraically if the polynomial is given in factorised form or if the degree is less than $5$5.
In the case of quadratic functions, the zeros, if there are any, are found by the 'completing the square' process, or, which amounts to the same thing, with the help of the quadratic formula. There are more complicated formulas for the roots of cubic and quartic polynomial equations.
The zeros of polynomial functions of degree $5$5 or more can not always be expressed in a precise algebraic form. However, their locations can be found approximately by trial-and-error or other numerical methods.
Having found the intercepts of the graph on the two axes, we study the behaviour of the function by looking for turning points or stationary points and by investigating the sign of the gradients near such points. For this, we use differentiation and the fact that the gradient is zero at a stationary point.
If the gradient before a stationary point is positive and negative after, the stationary point must be a local maximum. If the gradient changes from negative before to positive after a stationary point, the stationary point must be a minimum. If the gradient is the same on both sides of a stationary point, we say the stationary point is a point of inflection.
Find enough important points on the graph of the quadratic function $y$y defined on the domain $[-3,3]$[−3,3] by $y(x)=\frac{x^2}{3}-x+\frac{2}{3}$y(x)=x23−x+23, to enable sketching of the graph.
The intercept on the $y$y-axis is $y(0)=\frac{2}{3}$y(0)=23. We should also, at this stage, find the coordinates of the endpoints of the graph. These are $\left(-3,y(-3)\right)$(−3,y(−3)) and $\left(3,y(3)\right)$(3,y(3)). That is, $\left(-3,6\frac{2}{3}\right)$(−3,623) and $\left(3,\frac{2}{3}\right)$(3,23).
To find the zeros of $y$y, we solve $y(x)=0$y(x)=0. That is, $\frac{x^2}{3}-x+\frac{2}{3}=0$x23−x+23=0.
$0$0 | $=$= | $\frac{x^2}{3}-x+\frac{2}{3}$x23−x+23 |
$=$= | $x^2-3x+2$x2−3x+2 | |
$=$= | $(x-2)(x-1)$(x−2)(x−1) |
Therefore, the roots are $x=2$x=2 and $x=1$x=1. That is, the intercepts on the $x$x-axis are the points $(1,0)$(1,0) and $(2,0)$(2,0).
To find a turning point, we differentiate. Thus, $y'(x)=\frac{2x}{3}-1$y′(x)=2x3−1. Setting $y'(x)=0$y′(x)=0, we find $x=\frac{3}{2}$x=32. Then, to locate the point we find the function value $y\left(\frac{3}{2}\right)=-\frac{1}{12}$y(32)=−112. The coordinates are $\left(\frac{3}{2},-\frac{1}{12}\right)$(32,−112).
If $x<\frac{3}{2}$x<32, say $x=1$x=1, we have $y'(1)=-\frac{1}{3}$y′(1)=−13 which is negative. When $x>\frac{3}{2}$x>32, say $x=2$x=2, we have $y(2)=\frac{1}{3}$y(2)=13 which is positive. So, the stationary point is a local minimum.
The following graph includes the points found in the above discussion.
Consider the function $f\left(x\right)=9x^2+18x-16$f(x)=9x2+18x−16.
State the coordinates of the $y$y-intercept.
Give your answer in the form $\left(a,b\right)$(a,b).
Solve for the $x$x-value(s) of the $x$x-intercept(s).
If there is more than one value, write all of them on the same line, separated by commas.
Determine an equation for $f'\left(x\right)$f′(x).
Hence solve for the $x$x-coordinate(s) of the stationary point(s).
If there is more than one, write all of them on the same line separated by commas.
By completing the table of values, find the gradient of the curve for the following values of $x$x:
$x$x | $-2$−2 | $-1$−1 | $0$0 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Select the correct statement.
$\left(-1,-25\right)$(−1,−25) is a minimum turning point.
$\left(-1,-25\right)$(−1,−25) is a maximum turning point.
$\left(-1,-25\right)$(−1,−25) is a minimum turning point.
$\left(-1,-25\right)$(−1,−25) is a maximum turning point.
Draw the graph below.
Consider the function $f\left(x\right)=\left(4x+5\right)^2\left(x-1\right)$f(x)=(4x+5)2(x−1).
State the coordinates of the $y$y-intercept.
Give your answer in the form $\left(a,b\right)$(a,b).
Solve for the $x$x-value(s) of the $x$x-intercept(s).
If there is more than one value, write all of them on the same line, separated by commas.
Determine an equation for $f'\left(x\right)$f′(x).
Hence solve for the $x$x-coordinate(s) of the stationary point(s).
If there is more than one, write all of them on the same line separated by commas.
By completing the tables of values, find the gradient of the curve at $x=-2$x=−2, $x=-1$x=−1, $x=0$x=0 and $x=1$x=1.
$x$x | $-2$−2 | $-\frac{5}{4}$−54 | $-1$−1 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $0$0 | $\editable{}$ |
$x$x | $0$0 | $\frac{1}{4}$14 | $1$1 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $0$0 | $\editable{}$ |
Select the correct statement(s).
$\left(\frac{1}{4},-27\right)$(14,−27) is a maximum turning point.
$\left(-\frac{5}{4},0\right)$(−54,0) is a minimum turning point.
$\left(\frac{1}{4},-27\right)$(14,−27) is a minimum turning point.
$\left(-\frac{5}{4},0\right)$(−54,0) is a maximum turning point.
$\left(\frac{1}{4},-27\right)$(14,−27) is a maximum turning point.
$\left(-\frac{5}{4},0\right)$(−54,0) is a minimum turning point.
$\left(\frac{1}{4},-27\right)$(14,−27) is a minimum turning point.
$\left(-\frac{5}{4},0\right)$(−54,0) is a maximum turning point.
Draw the graph below.
Consider the function $f\left(x\right)=\left(x+2\right)^3-1$f(x)=(x+2)3−1.
State the coordinates of the $y$y-intercept.
Give your answer in the form $\left(a,b\right)$(a,b).
Solve for the $x$x-value(s) of the $x$x-intercept(s).
If there is more than one value, write all of them on the same line, separated by commas.
Determine an equation for $f'\left(x\right)$f′(x).
Hence solve for the $x$x-coordinate(s) of the stationary point(s).
If there is more than one, write all of them on the same line separated by commas.
By completing the tables of values, find the gradient of the curve at $x=-3$x=−3 and $x=-1$x=−1.
$x$x | $-3$−3 | $-2$−2 | $-1$−1 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $0$0 | $\editable{}$ |
Select the correct statement.
$\left(-2,-1\right)$(−2,−1) is a minimum turning point.
$\left(-2,-1\right)$(−2,−1) is a maximum turning point.
$\left(-2,-1\right)$(−2,−1) is horizontal point of inflection.
$\left(-2,-1\right)$(−2,−1) is a minimum turning point.
$\left(-2,-1\right)$(−2,−1) is a maximum turning point.
$\left(-2,-1\right)$(−2,−1) is horizontal point of inflection.
Draw the graph below.
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
Apply differentiation methods in solving problems