find the difference between the endpoints of the interval on the y-axis. (1) [y-value at x=p] - (2) [y-value at x=p+k]

I know you don't need this anymore but for the ones who do this is the answerLook at the graph and trust it. Yes, I know it says to substitute and you can following the last page on part a) Doing what they do in the tables and it will get you the same answer as if you'd just look at the graph.

The answer for the y- value at x=p IS 8. and the answer for the y-value at x=p+k IS 5. So go ahead and write that down if you haven't already!

For step2 Find the difference between the endpoints of the interval on the y-axis. The answer to your question, Is and always will be 3! The difference is 3 because you subtract 3 from 8 and you get 5.

If you didn't read clearer, I'll write it for you again, The length interval is K, it is the length between p and p+k.

For the following questions, does your answer to step 2 depend on p, well I suppose. This is whyThe answer does depend on p because we are dealing with a slope. if you move up or down the slope it will not be the same.

For the question Is the difference between the y-values the same regardless of where the interval starts? The answer is yes But WRITE YES! Why? Because if the interval moves as a whole the difference between the y-values is the same! Try it for yourself. Move K back two x-values and the difference for the y-values will be the same, IT WILL STILL BE THREE.

Hope you understood!

And for part one the answer on part c (-2). It depends on K.

b

step-by-step explanation:

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

ver más en -

step-by-step explanation: