find equation of parabola?

Question

if x intercepts are 1 and 5, and y intercept is 2

Answer 1

The basic form of the equation of a parabola is y = ax^2 + bx + c. The x-intercepts are the solutions to the quadratic equation (when y=0). We know that the equation will equal 0 when x is 1 or x is 5, so we can represent this in factored form as:

a(x-1)(x-5) = 0, where a is just some constant. (a != 0). Notice that if you plug in 1 or 5 for x, this equation holds true, regardless of what a is.

But now we need to figure out what the equation is when y is NOT zero. So for any x, our basic equation is going to be:
y = a(x-1)(x-5)
We need to figure out what a is, so let's use that last piece of info: y-intercept is 2. This basically means we have the point (0,2) that we can plug into our equation to solve for a:
2 = a(0-1)(0-5)
2 = a(-1)(-5)
2 = 5a
2/5 = a

So now that we know what a is, we have our final equation:
y = (2/5)(x-1)(x-5) OR if you FOIL it out:
y = (2/5)x^2 - (12/5)x + 2

Answer 2

Start with a second degree equation that's close.

y = (x - 1)(x - 5)

This clearly has zeros at x = 1 and 5. But when x = 0, y = 5.
To make the y intercept 2, multiply by 2/5.

y = (2/5)(x - 1)(x - 5)

If you want this in more traditional form multiply it out.

y = (2/5)(x² - 6x + 5) = (2/5)(x² - 6x + 9 - 4)
y = (2/5)(x² - 6x + 9) - (2/5)(4)
y = (2/5)(x - 3)² - 8/5