Math HELP! plz can you help me solve these?

Question

1)Find the equation of a parabola with vertex (3,4) and y-intercept (0-5)

2) The sum of two the numbers 325 and the difference of the squares of these two numbers are 125. Use the system of two equations to find the numbers.

3) Write y=1/4X^2 +4X -7 in the form y= a(x-h)^2 +k

Answer 1

(1)
h = 3
k = 4
5 = a(0-3)² + 4
5 = 9a + 4
1 = 9a
(1/9) = a

Y = (1/9)(x - 3)² + 4

(3)
Y = (¼)x² + 4x - 7
Y = (¼)(x²+16x) - 7

x² +16x = x² + 16x + (16/2)² - (16/2)²
= [x + (16/2)]² - (16/2)
= (x + 8)² - 8²
= (x + 8)² - 64

Y = (¼)[(x + 8)² - 64] - 7
Y = (¼)(x + 8)² - 16 - 7
Y = (¼)(x + 8)² - 23

Answer 2

1)Find the equation of a parabola with vertex (3,4) and y-intercept (0-5)
since the vertex (or in this case, the center) of the parabola is at (3,4), we know that the equation must take the form of either of the following:
(x-3)^2 = 4p(y-4) or (y-4)^2 = 4p(x-3)
since we dont know, plut the y-intercept points into both equations to find p
(0-3)^2 = 4p(-5-4) or (-5-4)^2 = 4p(0-3)
9 = 4p(-9) or 81 = 4p(-3)
-1 = 4p or -27 = 4p
thus, the equation of the parabola could be either
(x-3)^2 = -1(y-4) or (y-4)^2 = -27(x-3)

2) The sum of two the numbers 325 and the difference of the squares of these two numbers are 125. Use the system of two equations to find the numbers.
set up a system of equations:
x + y = 325 y = 325 - x
x^2 - y^x = 125
plug the first equation into the second to get the value of x
x^2 - (325-x)^2 = 125
x^2 - (105625 - 650x + x^2) = 125
distribute the negative
x^2 - 105625 + 650x - x^2) = 125
the x^2 cancel each other out and add both sides by 105625
650x = 105750
divide both sides by 650
x = 2115/13
plug x back into the first equation to get y
x + y = 325
2115/13 + y = 325
y = 2110/13

thus the answers for x and y are:
x = 2115/13
y = 2110/13

3) Write y=1/4X^2 +4X -7 in the form y= a(x-h)^2 +k
add 7 to both sides
7 + y = (1/4)x^2 + 4x
complete the square
7 + y + (1/16) = 1/4(x^2 + x + 1/4)
(113/16) + y = 1/4(x + (1/2))^2
y = 1/4(x + (1/2))^2 - (113/16)

Answer 3

You have some fairly decent problems here. Usually I don't like to just give answers, but I will see what I can do. How about the second one?

a + b = 325 (sum two numbers)

a^2 - b^2 = 125 (we will assume "a" is the larger number - it won't hurt the problem)

Use substitution: Since a + b = 325, then b = 325 - a

and substitute that into the other equation:

a^2 - (325-a)^2 = 125

FOIL out:

a^2 - (105625 - 650a + a^2) = 125

simplify by distributing the negative:

a^2 - 105625 + 650a - a^2 = 125

simplify more:

650a = 105750

a = 2115 / 13

b = 325 - a

b = 2110 / 13

Check it...see if it's right.

I hope I haven't ruined your schooling by helping you on one little problem.

The first one is actually very very easy. Start with the vertex form of a parabola:

y= a(x-h)^2 +k

You know the vertex of a parabola is located at (h, k), so:

h = 3
k = 4

Plug those into that equation. You also know a generic point on the parabola is (x, y), and since they give you a generic point (0, 5), so:

x = 0
y = 5 (or negative 5 if you made a mistake in writing the problem).

Plug x, y, h, k into y= a(x-h)^2 +k, find "a", then rewrite the whole thing leaving x and y as your variables.

Answer 4

errr...dont feel like doing the first one (just do some of that -b/2a and plug back in)

The second one:

x+y=325
x^2-y^2=125

then solve it...