Writing equations for parabolas?

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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...Show more

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...Show more

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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 ...Show more