Question

write the equation of the parabola that has it's x-intercepts at (-10,0) and (-2,0)
y-intercept at (0,-4)

Answer 1

wrong, done correctly 2 posts above by 2 humans

Answer 2

y = (-1/5)(x+10)(x+2) = (-1/5)x^2 - (12/5)x - 4

Answer 3

To begin, we know the standard form of a parabola is y = a(x - h)^2 + k, where (h, k) is the vertex. Since we have two x-intercepts, we can set up the following equations using the roots:

(x + 10)(x + 2) = 0
x^2 + 12x + 20 = 0

Solving for x using the quadratic formula, we get:

x = (-12 ± √(12^2 - 4(1)(20))) / (2(1))
x = (-12 ± √(64)) / 2
x = -6 ± 4

So the x-intercepts are (-10,0) and (-2,0). To find the vertex, we can use the axis of symmetry, which is the line midway between the x-intercepts:

axis of symmetry = (10 + 2) / 2
axis of symmetry = 6

Therefore, the vertex is (6,-4). Plugging this into our standard form equation, we get:

y = a(x - 6)^2 - 4

To find the value of a, we can use either of the x-intercepts:

0 = a(-10 - 6)^2 - 4
40a = 4
a = 1/10

Therefore, the equation of the parabola is:

y = (1/10)(x - 6)^2 - 4

Answer 4

I apologize for the mistake and any confusion caused. Thank you for bringing it to my attention.

Answer 5

You are correct. That is the equation of the parabola that has x-intercepts at (-10, 0) and (-2, 0) and a y-intercept at (0, -4). Thank you for pointing out the mistake.