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Write the equation of the parabola that has its x-intercepts

at (5, 0) and (-6, 0) and its
at (0, -1).

User Jbastos
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2 Answers

1 vote

5/0-6/0-41=

hope that helps okay :) if not really really sorry

User Pranay Soni
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8.9k points
5 votes

Answer:

y = (1/25)x^2 - 1

Explanation:

To begin, we can use the fact that the x-intercepts of the parabola are at (5, 0) and (-6, 0) to determine the x-intercepts form of the equation. The x-intercepts form of a parabola is given by:

(x - r)(x - s) = 0,

where r and s are the x-coordinates of the x-intercepts.

Using this form, we can write:

(x - 5)(x + 6) = 0

Expanding this equation gives:

x^2 + x - 30 = 0

Next, we use the fact that the vertex of the parabola is at (0, -1) to determine the y-intercept form of the equation. The y-intercept form of a parabola is given by:

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

where (h, k) is the vertex.

Substituting (0, -1) into this equation gives:

y = a(x - 0)^2 - 1

Simplifying gives:

y = ax^2 - 1

Now we can use the fact that the parabola passes through the point (5, 30) to determine the value of a. Substituting x = 5 and y = 0 into the equation above gives:

0 = a(5)^2 - 1

Solving for a gives:

a = 1/25

Substituting this value of a into the equation above gives the final equation of the parabola:

y = (1/25)x^2 - 1

User MD Jahid Hasan
by
7.9k points