However,
[link] is only valid if
$-\frac{q}{a}>0$ which means that either
$q<0$ or
$a<0$ but not both. This is consistent with what we expect, since if
$q>0$ and
$a>0$ then
$-\frac{q}{a}$ is negative and in this case the graph lies above the
$x$ -axis and therefore does not intersect the
$x$ -axis. If however,
$q>0$ and
$a<0$ , then
$-\frac{q}{a}$ is positive and the graph is hat shaped with turning point above the
$x$ -axis and should have two
$x$ -intercepts. Similarly, if
$q<0$ and
$a>0$ then
$-\frac{q}{a}$ is also positive, and the graph should intersect with the
$x$ -axis twice.
For example, the
$x$ -intercepts of
$g\left(x\right)={(x-1)}^{2}+2$ are given by setting
$y=0$ to get:
which has no real solutions. Therefore, the graph of
$g\left(x\right)={(x-1)}^{2}+2$ does not have any
$x$ -intercepts.
Intercepts
Find the x- and y-intercepts of the function
$f\left(x\right)={(x-4)}^{2}-1$ .
Find the intercepts with both axes of the graph of
$f\left(x\right)={x}^{2}-6x+8$ .
Given:
$f\left(x\right)=-{x}^{2}+4x-3$ . Calculate the x- and y-intercepts of the graph of
$f$ .
Turning points
The turning point of the function of the form
$f\left(x\right)=a{(x+p)}^{2}+q$ is given by examining the range of the function. We know that if
$a>0$ then the range of
$f\left(x\right)=a{(x+p)}^{2}+q$ is
$\left\{f\right(x):f(x)\in [q,\infty \left)\right\}$ and if
$a<0$ then the range of
$f\left(x\right)=a{(x+p)}^{2}+q$ is
$\left\{f\right(x):f(x)\in (-\infty ,q\left]\right\}$ .
So, if
$a>0$ , then the lowest value that
$f\left(x\right)$ can take on is
$q$ . Solving for the value of
$x$ at which
$f\left(x\right)=q$ gives:
$\therefore $$x=-p$ at
$f\left(x\right)=q$ . The co-ordinates of the (minimal) turning point is therefore
$(-p,q)$ .
Similarly, if
$a<0$ , then the highest value that
$f\left(x\right)$ can take on is
$q$ and the co-ordinates of the (maximal) turning point is
$(-p,q)$ .
Turning points
Determine the turning point of the graph of
$f\left(x\right)={x}^{2}-6x+8$ .
Given:
$f\left(x\right)=-{x}^{2}+4x-3$ . Calculate the co-ordinates of the turning point of
$f$ .
Find the turning point of the following function by completing the square:
$y=\frac{1}{2}{(x+2)}^{2}-1$ .
Axes of symmetry
There is only one axis of symmetry for the function of the form
$f\left(x\right)=a{(x+p)}^{2}+q$ . This axis passes through the turning point and is parallel to the
$y$ -axis. Since the
$x$ -coordinate of the turning point is
$x=-p$ , this is the axis of symmetry.
Axes of symmetry
Find the equation of the axis of symmetry of the graph
$y=2{x}^{2}-5x-18$ .
Write down the equation of the axis of symmetry of the graph of
$y=3{(x-2)}^{2}+1$ .
Write down an example of an equation of a parabola where the y-axis is the axis of symmetry.
Sketching graphs of the form
$f\left(x\right)=a{(x+p)}^{2}+q$
In order to sketch graphs of the form
$f\left(x\right)=a{(x+p)}^{2}+q$ , we need to determine five characteristics:
sign of
$a$
domain and range
turning point
$y$ -intercept
$x$ -intercept
For example, sketch the graph of
$g\left(x\right)=-\frac{1}{2}{(x+1)}^{2}-3$ . Mark the intercepts, turning point and axis of symmetry.
Firstly, we determine that
$a<0$ . This means that the graph will have a maximal turning point.
The domain of the graph is
$\{x:x\in \mathbb{R}\}$ because
$f\left(x\right)$ is defined for all
$x\in \mathbb{R}$ . The range of the graph is determined as follows:
Therefore the range of the graph is
$\left\{f\right(x):f(x)\in (-\infty ,-3\left]\right\}$ .
Using the fact that the maximum value that
$f\left(x\right)$ achieves is -3, then the
$y$ -coordinate of the turning point is -3. The
$x$ -coordinate is determined as follows:
which has no real solutions. Therefore, there are no
$x$ -intercepts.
We also know that the axis of symmetry is parallel to the
$y$ -axis and passes through the turning point.
Sketching the parabola
Draw the graph of
$y=3{(x-2)}^{2}+1$ showing all the intercepts with the axes as well as the coordinates of the turning point.
Draw a neat sketch graph of the function defined by
$y=a{x}^{2}+bx+c$ if
$a>0$ ;
$b<0$ ;
${b}^{2}=4ac$ .
Writing an equation of a shifted parabola
Given a parabola with equation
$y={x}^{2}-2x-3$ . The graph of the parabola is shifted one unit to the right. Or else the y-axis shifts one unit to the left i.e.
$x$ becomes
$x-1$ . Therefore the new equation will become:
Show that if
$a<0$ , then the range of
$f\left(x\right)=a{(x+p)}^{2}+q$ is
$\left\{f\right(x):f(x)\in (-\infty ,q\left]\right\}$ .
If (2,7) is the turning point of
$f\left(x\right)=-2{x}^{2}-4ax+k$ , find the values of the constants
$a$ and
$k$ .
The graph in the figure is represented by the equation
$f\left(x\right)=a{x}^{2}+bx$ . The coordinates of the turning point are (3,9). Show that
$a=-1$ and
$b=6$ .
Given:
$y={x}^{2}-2x+3$ . Give the equation of the new graph originating if:
The graph of
$f$ is moved three units to the left.
The
$x$ -axis is moved down three units.
A parabola with turning point (-1,-4) is shifted vertically by 4 units upwards. What are the coordinates of the turning point of the shifted parabola?
Questions & Answers
where we get a research paper on Nano chemistry....?
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
How this robot is carried to required site of body cell.?
what will be the carrier material and how can be detected that correct delivery of drug is done
Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?