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Writing equations for parabolas?
I have to write the equation for a parabola with a y intercept of 10, x intercept of 2, and equation of axis of symmetry of x-3=0. Any help?
3 Answers
- stanschimLv 79 years agoFavorite Answer
Assume that the parabola is in standard form: y = ax^2 + bx + c.
(0,10) is on the parabola implies: 10 = a(0^2) + b(0) + c, or c = 10
(2,0) is on the parabola implies: 0= a(2^2) + 2b + c, or 0 = 4a + 2b + 10, or 2a + b = -5
We also know the axis of symmetry runs through an x-coordinate of -b/2a = 3, or b = -6a
2a + b = -5
2a - 6a = -5
-4a = -5
a = 5/4
2(5/4) + b = -5
(5/2) + b = -5
b = -5 - (5/2) = (-10/2) - (5/2) = -15/2
y = (5/4)x^2 - (15/2)x + 10 <== answer
- Josh KLv 69 years ago
Line of symmetry is x = 3, and an x intercept is 2, then another x intercept must be x = 4
3-1 and 3+1
Then y = c(x-2)(x-4)
Now sub in the y intercept of (0,10) to find c
10 = c*-2*-4 = 8c
c = 5/4
y = 5/4(x-2)(x-4)
I hope this helps
- Wile E.Lv 79 years ago
x1 = 2
h = 3
y int. = 10, so
c = 10
x2 = h + (h - x1)
x2 = 3 + (3 - 2)
x2 = 3 + 1
x2 = 4
Roots: (2, 0) & (4, 0)
y = a (x - 2)(x - 4)
y = a (x² - 6x + 8)
y = ax² - 6ax + 8a
Since c = 10 and
8a = c, then
8a = 10, so
a = 10 / 8
a = 5/4
Equation:
y = 5/4 x² - 6(5/4)x + 8(5/4)
y = 5/4 x² - 30/4 x + 40/4
y = 5/4 x² - 15/2 x + 10
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Source(s): 4/14/12