Question

How would I write an equation of quadratic equation with zeros of (–2, 0) and (5, 0), and passes through the point: (7, –4)

Answer 1

Nice question.

Well, remember (or not :-) ) that a quadratic equation with roots a and b can always be factorized in the form

(x-a)(x-b)

because if (x-a)(x-b) = 0, then x = a or x = b

But notice this isn't the only quadratic that has roots a and b.

So does 2(x-a)(x-b) and -5(x-a)(x-b)

Hence in general for any c not equal to zero the quadratic

c(x-a)(x-b)

has roots at x = a or b.

That's the theory.

Now for your problem you are given these two points on the graph: (-2,0) and (5,0) and told they are zeros, i.e., they are roots.

So the GENERAL form of the quadratic with those roots is

y = c(x+2)(x-5) --- **

Now you need find the value for c which makes the graph of this quadratic pass through (7,-4).

Hence substitute into the equation ** above x = 7 and y = -4 and solve for c.

Answer 2

So like this: -4=c(7+2)(7-5)
and then solve for c?

Answer 3

Thank you so much!

Answer 4

I got y=(-2/9)(x+2)(x-5) as the answer

Answer 5

I'd expand it out, as that's the usual form, but I think that looks right.