We then substitute this into the equation of the graph (i.e.
$y={x}^{2}$ ) to determine the
$y$ -coordinate of the turning point:
$f\left(0\right)={\left(0\right)}^{2}=0$ This corresponds to the point that we have previously calculated.
Calculate the turning points of the graph of the function
$f\left(x\right)=2{x}^{3}-9{x}^{2}+12x-15$ .
Using the rules of differentiation we get:
${f}^{\text{'}}\left(x\right)=6{x}^{2}-18x+12$
The turning points of the graph of
$f\left(x\right)=2{x}^{3}-9{x}^{2}+12x-15$ are (2,-11) and (1,-10).
We are now ready to sketch graphs of functions.
Method:
Sketching Graphs:
Suppose we are given that
$f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d$ , then there are
five steps to be followed to sketch the graph of the function:
If
$a>0$ , then the graph is increasing from left to right, and may have a maximum and a minimum. As
$x$ increases, so does
$f\left(x\right)$ . If
$a<0$ , then the graph decreasing is from left to right, and has first a minimum and then a maximum.
$f\left(x\right)$ decreases as
$x$ increases.
Determine the value of the
$y$ -intercept by substituting
$x=0$ into
$f\left(x\right)$
Determine the
$x$ -intercepts by factorising
$a{x}^{3}+b{x}^{2}+cx+d=0$ and solving for
$x$ . First try to eliminate constant common factors, and to group like terms together so that the expression is expressed as economically as possible. Use the factor theorem if necessary.
Find the turning points of the function by working out the derivative
$\frac{df}{dx}$ and setting it to zero, and solving for
$x$ .
Determine the
$y$ -coordinates of the turning points by substituting the
$x$ values obtained in the previous step, into the expression for
$f\left(x\right)$ .
Use the information you're given to plot the points and get a rough idea of the gradients between points. Then fill in the missing parts of the function in a smooth, continuous curve.
Draw the graph of
$g\left(x\right)={x}^{2}-x+2$
The
$y$ -intercept is obtained by setting
$x=0$ .
$g\left(0\right)={\left(0\right)}^{2}-0+2=2$ The turning point is at (0,2).
The
$x$ -intercepts are found by setting
$g\left(x\right)=0$ .
Using the quadratic formula and looking at
${b}^{2}-4ac$ we can see that this would be negative and so this function does not have real roots. Therefore, the graph of
$g\left(x\right)$ does not have any
$x$ -intercepts.
Work out the derivative
$\frac{dg}{dx}$ and set it to zero to for the
$x$ coordinate of the turning point.
$\frac{dg}{dx}=2x-1$
If we divide
$g\left(x\right)$ by
$(x-1)$ we are left with:
$-{x}^{2}+5x-4$ This has factors
$-(x-4)(x-1)$
Therefore:
$g\left(x\right)=-(x-1)(x-1)(x-4)$
The
$x$ -intercepts are:
${x}_{int}=1,4$
Sketching graphs
Given
$f\left(x\right)={x}^{3}+{x}^{2}-5x+3$ :
Show that
$(x-1)$ is a factor of
$f\left(x\right)$ and hence fatorise
$f\left(x\right)$ fully.
Find the coordinates of the intercepts with the axes and the turning points and sketch the graph
Sketch the graph of
$f\left(x\right)={x}^{3}-4{x}^{2}-11x+30$ showing all the relative turning points and intercepts with the axes.
Sketch the graph of
$f\left(x\right)={x}^{3}-9{x}^{2}+24x-20$ , showing all intercepts with the axes and turning points.
Find the equation of the tangent to
$f\left(x\right)$ at
$x=4$ .
Local minimum, local maximum and point of inflextion
If the derivative (
$\frac{dy}{dx}$ ) is zero at a point, the gradient of the tangent at that point is zero. It means that a turning point occurs as seen in the previous example.
From the drawing the point (1;0) represents a
local minimum and the point (3;4) the
local maximum .
A graph has a horizontal
point of inflexion where the derivative is zero but the sign of the sign of the gradient does not change. That means the graph always increases or always decreases.
From this drawing, the point (3;1) is a horizontal point of inflexion, because the sign of the derivative does not change from positive to negative.
Questions & Answers
Is there any normative that regulates the use of silver nanoparticles?
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?