a parabola has x-intercepts at x=1 and x=-3. it also passes through the points (0,-6). find the equation

The general form of a quadratic equation for a parabola is:

\[
y = a(x - r_1)(x - r_2)
\]

Where \( r_1 \) and \( r_2 \) are the x-intercepts. Given that the x-intercepts are \( x = 1 \) and \( x = -3 \), we can substitute these values for \( r_1 \) and \( r_2 \), giving:

\[
y = a(x - 1)(x + 3)
\]

Now, to find the value of \( a \), we use the fact that the parabola passes through the point \( (0, -6) \). Substituting \( x = 0 \) and \( y = -6 \) into the equation:

\[
-6 = a(0 - 1)(0 + 3)
\]
\[
-6 = a(-1)(3)
\]
\[
-6 = -3a
\]
\[
a = 2
\]

Thus, the equation of the parabola is:

\[
y = 2(x - 1)(x + 3)
\]

Expanding this equation:

\[
y = 2(x^2 + 3x - x - 3)
\]
\[
y = 2(x^2 + 2x - 3)
\]
\[
y = 2x^2 + 4x - 6
\]

So, the equation of the parabola is:

\[
y = 2x^2 + 4x - 6
\]

Would you like further details or have any questions? 

Here are some related questions for further exploration:
1. How do you find the vertex of this parabola?
2. What is the axis of symmetry of the parabola?
3. How do you determine if the parabola opens upwards or downwards?
4. How do x-intercepts help in determining the shape of the parabola?
5. Can this equation be rewritten in vertex form?

**Tip:** The coefficient \(a\) controls how "wide" or "narrow" the parabola is, and whether it opens upwards (positive \(a\)) or downwards (negative \(a\)).

A parabola has x-intercepts at x=1 and x=-3. It also passes through the point (0,-6). Find the equation.

Solve for the equation of a parabola with x-intercepts at x=1 and x=-3, passing through the point (0,-6). Explore algebraic concepts like quadratic equations and the intercept form, suitable for Grades 9-10.

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