In graphing a trigonometric function, how does one establish which are the EXTREMUM coordinates and which are the MIDLINE INTERCEPTION coordinates. I can find the x values and the y values but do not know which X goes with which Y and the graphs end up incorrect! It is EXTREMELY frustrating. How do I discern which is which Thank You.

If I getting the question correctly, your doubt is the order of the values you get after doing the math. I mean, you can do the calculations, but in the end, you don't form the coordinate pairs correctly, that's why I understand from your question. So, here is what you need to do.

One way to graph is by using the definitions of those kinds of functions. For example, let's say we want to find the points to draw the function:

Remember that trigonometric functions have a specific period, that means, their drawing repeats over and over again after a certain number. That period is , that means this number represents a cycle.

So, the main thing you need to do is to pick a starting point to then, draw the curve according to the period.

Now, we know that sine functions intercept the origin of the coordinate system, as you can observe in the image attached, there you can see that from the origin you draw those waves making sure you intercept the x-axis at every number. In the end, you will have a sine function.

On the other hand, if you want to have a chart with all x-values and y-values. First, you need to set x-values: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. Then, you need to find each y-values for each of them.

Now, you have to draw a chart value, to keep in other the coordinate pairs, that way, you'll have the correct pairs at the end.

For example, for , we get , that means in the chart value, you are gonna form the pair , and now you have your first point of the drawing. Then, you keep repeating the process until you complete all the chart values.

As you can imagine, you're going to get really small decimal numbers, that's why I explained to you the first method, it's faster and easier.

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: