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write an equation of the parabola that passes through the point (-1,-3) and has x intercepts of -2 and 4

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The vertex form of a parabola is:

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

where (h, k) is the vertex of the parabola and "a" is a constant that determines the shape of the parabola.

First, we can find the vertex of the parabola using the x-coordinate of the midpoint of the x-intercepts:

Midpoint of x-intercepts = (-2 + 4)/2 = 1

Therefore, the x-coordinate of the vertex is 1.

Since the parabola passes through the point (-1, -3), we can substitute these values into the vertex form to get:

-3 = a(-1 - 1)^2 + k
-3 = 4a + k

We also know that the parabola has x-intercepts at -2 and 4, which means that the parabola intersects the x-axis at these points. This tells us that the vertex is in the middle of these two points, or:

(1,0) = (-2 + 4)/2

We can plug this information into the vertex form to find the value of "a":

0 = a(1 - 1)^2 + k
0 = k

This means that the vertex is at (1,0), and the equation of the parabola is:

y = a(x - 1)^2

To find the value of "a", we can use the fact that the parabola passes through the point (-1,-3):

-3 = a(-1 - 1)^2
-3 = 16a
a = -3/16

Substituting this value of "a" into the vertex form, the equation of the parabola is:

y = -(3/16)(x - 1)^2
User Gerald Chifanzwa
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