The zeros of a parabola are 6 and −5. if (-1, 3) is a point on the graph, which equation can be solved to find the value of a in the equation of the parabola?

y = (3/32)(x^2 - x - 30)

Step-by-step explanation:

The equation of this parabola looks like y = (x -  6)(x + 5).

It is possible that there's a constant coefficient:  y = a(x -  6)(x + 5).

Multiplying out y = (x -  6)(x + 5), we get y = x^2 - x - 30.

Does (-1, 3) satisfy this equation?    3 = (-1)^2 - (-1) - 30    NO, this is not true.

So, use the equation y = a(x -  6)(x + 5)  instead.

This is equivalent to y = a(x^2 - x - 30).  Determine the value of a that makes this equation true for the point (-1, 3):

3 = a[(-1)^2 + 1 + 30], or

3 = a[32].  Thus, a must be 3/32, and the equation is

y = (3/32)(x^2 - x - 30)

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m a t h w a y. c o m

step-by-step explanation: