Write an equation for the parabola that passes through (4, 3) and has x-intercepts of −1 and 5

Question

Write an equation for the parabola that passes through (4, 3) and has x-intercepts of  and -1 and 5

Answer 1

If the requested parabola has x-intercepts of -1 and 5, then that means that when y=0, x=-1 and x=5.

If x=-1, then add 1 to each side to get that x+1=0. This is our first factor. Similarly, if x=5, subtract 5 from each side to get x-5=0. This is the other factor. A quadratic is two factors (not necessarily unique) with a nonzero coefficient in front. So our quadratic will look something like

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

where a is some number.

How do we find a? We are told that this quadratic passes through the point (4, 3). That is, when x=4, y=3. Plug those in and solve for a.

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

3=a(4+1)(4-5)

3=a(5)(-1)

3=-5a

3/-5=a

-3/5=a

Thus,

y=-3/5(x+1)(x-5)

You might want to foil it out so it is in standard form, ax^2+bx+c.

Answer 2

f(x) = A(x + 1)(x - 5)

3 = A(4 + 1)(4 - 5)

3 = -5A

A = -3/5

f(x) = (-3/5)(x + 1)(x - 5)