graph d=1.5+0.8t
show axial intercepts and any turning points
how would i do this? plug in a number for t or? and how would i find x and y and graph it
the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points.
check out the intersect function.
here is one approach:
% generate a rectangular boundry
rectx = [0.1 0.1 0.9 0.9];
recty = [0.1 0.5 0.5 0.1];
rect = polyshape(rectx,recty);

% initialize a figure for display purposes
figure(1);clf
titles = [...
 "all lines";...
 "partially within";...
 "exclusively within"];
for i = 1:3
 subplot(1,3,i)
 hold on
 title(titles(i))
 plot(rect)
 xlim([0 1])
 ylim([0 1])
end

% define the number of line segments to generate
nlines = 25;

% initialize a structure to store the lines (if you need the data later)
lines = struct();

% generate each line, calculate it's intersection conditions, and plot
for i=1:nlines
 
 % generate points for a random line segment
 lines(i).points = rand(2);
 
 % section the line into parts at intersection with the rectangle
 [in, out] = intersect(rect, lines(i).points);
 
 % plot all the lines that were generated
 subplot(1,3,1)
 line(lines(i).points(:,1),lines(i).points(:,2))
 
 % plot lines that are partially within the boundary
 if isempty(in)==0
 subplot(1,3,2)
 line(lines(i).points(:,1),lines(i).points(:,2))
 end
 
 % plot lines that are exclusively within the boundary
 if isempty(out)==1
 subplot(1,3,3)
 line(lines(i).points(:,1),lines(i).points(:,2))
 end
end
here is the output: 1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 } there are websites that give you technical analysis indicators (including golden crosses and death crosses) such as

https://www.stockmonitor.com/stock-screener/death-cross-50ma-cross-down-200ma/

or you can simply graph the moving averages yourself.

i am not a big fan of technical analysis indicators so i have never used them. if you do not need the "wave" contents for a further use, just drop out (k) from all its entries.
if you need it for a further use, initialize it as 
wave=zeros(3,numel(t));
before the loop and address it as wave(k,:) within the loop.
commands "stem" are definitely incorrect. if you want to see plots of the dependences on t, they should look something like stem(t,combined) or plot(t,combined), etc. however, the t range in the second version (0,3) is much wider than the one of the first version (0,0.01): consequently, such a graph will show much (hugely) more oscillations--you will probably see nothing but a unifromly painted square.
i do not see any need in the "nold on" and "hold off" commands in the both version. ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow :roll\_eyes:

[[image]](https://aws1.discourse-cdn.com/business6/uploads/dynamobim/original/3x/6/3/6393f7d9277c8a0ff2b48f07b1e2352f69f463b0.png "image")

t1 = [9, 9, 99, 9, 6, 7, 8];

t9 = dscore.list.chop(t1, 3);

a = t9;

t2 = get.valueatindex(a, 0);

t4 = [3, 2];

t10 = dscore.list.ofrepeateditem(t4, 3);

a1 = t10;

t3 = get.valueatindex(a1, 0);

t11 = t2<1l> + t3<1l>;

a2 = t9;

t5 = get.valueatindex(a2, 1);

a3 = t10;

t6 = get.valuea… May as well promote my comment to an answer.


You need to add points to a `Curve2D` object, then adjust the control points of the curve, and then finally draw the curve. So here's some simple, almost pseudo code:



```
var array_of_line_points # This already has the vectors which describe our line
for point in array_of_line_points:
  # The "get_perpendicular_vector()" function returns a vector that's a copy of the point, yet has been slid along a line parallel to two neighboring points. The "distance" variable is how far the control point should be from the originating point.
  var control_point1 = get_perpendicular_vector(point, distance)
  var control_point2 = get_perpendicular_vector(point, -distance)
  curve.add_point(point, control_point1, control_point2)
# ...In the draw function
func _draw():
  draw_polyline(curve.get_baked_points(), red, 2.0)

```

I got the inspiration for this from [Rob Spencer's page on Spline Interpolation](http://scaledinnovation.com/analytics/splines/aboutSplines.html). I'd like to make this more thorough (i.e. defining the `get_perpendicular_vector()` function). But this is the best I can currently do. (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. Here's the diagram you requested. All the best!  

Greg  

![I need a serpentine belt routing diagram for a - c528ac5.gif](http://i.fixya.net/uploads/images/c528ac5.gif) this is not perfect but i hope it shows you one possible approach. i hand calculated the position of the whisker of each box plot and use that to place the y position of the labels.

plot<-q + scale\_x\_discrete(breaks=c("b","h"), labels=c("bs", "ah"))

plot2<-plot + theme(axis.title.x = element\_blank(…