The height y (in feet) of a ball thrown by a child is
y=−1/14x^2+4x+5
where x is the horizontal distance in feet from the point at which the ball is thrown.

(a) how high is the ball when it leaves the child's hand?


(b) what is the maximum height of the ball?


(c) how far from the child does the ball strike the ground?

  (a)  5 ft

  (b)  21 ft

  (c)  17.165 ft

Step-by-step explanation:

(a) When y = 0, y = ...

  y = -1/4·0² +4·0 +5

  y = 5

The ball is 5 ft high when it leaves the child's hand.

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(b) For quadratic ax²+bx+c, the x-coordinate of the vertex is ...

  x = -b/(2a)

For this height function, we have a=-1/4 and b = 4, so the x-coordinate of the vertex is ...

  x = (-4)/(2(-1/4)) = 8

The height when x = 8 is ...

  y = (-1/4·8 +4)·8 +5 = 2·8 +5 = 21

The maximum height of the ball is 21 feet.

__

(c) The ball will hit the ground where y=0, so we want find x at that point.

  0 = -1/4x² +4x +5

  0 = -1/4(x² -16x) +5 = -1/4(x² -16x +64) +21 . . . . complete the square

  0 = -1/4(x -8)² +21 . . . . . . . write as a square

  84 = (x -8)² . . . . . . . . . . . .  multiply by -4, add 84

  2√21 = x -8 . . . . . . . . . . . take the positive square root

  x = 8 + 2√21 ≈ 17.165 . . . . feet

The ball strikes the ground about 17.165 feet from the child.

a line is an defined derm because it

a)contains an infinite number of points b) can be used to create other geometric shapes c)describes something that does not have width or depth d) is a term that does not have a formal definition. a line is an undefined term because it: contains an infinite number of points.

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: