I thought this was already done. The zeroes mean
y = a(x-2)(x-15)
the point means
a(6-2)(6-15) = -5
a = 5/36
y = 5/36 (x-2)(x-15)
In standard form, that is (x-h)^2 = 4p(y-k) we get
(x - 17/2)^2 = 36/5 (y + 845/144)
Find the standard equation of a parabola with zeros at x=2 and x=15 passing through the point (6,−5).
thank you
Remember to write your answers as exact numbers (fractions), rather than a decimal approximation.
i need the answer in this form:
y=...(x+-)^2 +-.....
5 answers
Your equation will have the form
y = a(x-2)(x-15)
but (6,-5) is on it, so that lets us find a
-5 = a(6-2)(6-15)
a = 5/36
y = (5/36)(x^2 - 17x + 30)
= (5/36)(x^2 - 17/2 + 289/4 - 289/4 + 30)
= (5/36)( (x - 17/2)^2 - 289/4 + 30)
= (5/36)(x - 17/2)^2 - 845/144
y = a(x-2)(x-15)
but (6,-5) is on it, so that lets us find a
-5 = a(6-2)(6-15)
a = 5/36
y = (5/36)(x^2 - 17x + 30)
= (5/36)(x^2 - 17/2 + 289/4 - 289/4 + 30)
= (5/36)( (x - 17/2)^2 - 289/4 + 30)
= (5/36)(x - 17/2)^2 - 845/144
Sorry,I don't get it
thank you
thank you
never mind....I got it oobleck
thank you
thank you
passes through:
(2 , 0)
(15 , 0)
(6 , -5 )
y = a x^2 + bx + c
0 = 4 a +2 b + c
0 = 225 a + 15 b + c
-5 = 36 a + 6 b + c
3 equations, 3 unkmowns
221a +13 b = 0 .....so b = -221 a/13
189 a + 9 b= 5
so
189 a + 9 (-221 a/13) = 5
2457 a - 1989 a = 65
468 a = 65
a = 65/468
go back and get b and c
(2 , 0)
(15 , 0)
(6 , -5 )
y = a x^2 + bx + c
0 = 4 a +2 b + c
0 = 225 a + 15 b + c
-5 = 36 a + 6 b + c
3 equations, 3 unkmowns
221a +13 b = 0 .....so b = -221 a/13
189 a + 9 b= 5
so
189 a + 9 (-221 a/13) = 5
2457 a - 1989 a = 65
468 a = 65
a = 65/468
go back and get b and c