Find the equation in standard form?

Find the equation in standard form (y = ax^2 + bx + c) of the quadratic whose roots are 2 and 7 which passes through the point (5, -18). (also need: y= a(x-r1)(x-r2))

Find the equation in standard form?
If a quadratic has roots 2 and 7, then the equation of said quadratic will look like:


y = a(x-2)(x-7) (r1 and r2 represent root 1 and root 2)


Now plug in the point (5,-18) to solve for a:


-18 = a(5-2)(5-7)


I'll let you finish that.





Once you have a, simply FOIL the right hand side of the first equation I gave you and you'll have an equation in standard form. I'll leave that to you as well.
Reply:So first you have roots 2 and 7 which means


x = 2 and x = 7 or


x-2 = 0


x-7 = 0


now you also have a pt. (5, -18). 5 is x -18 is y


plug it into y = a (x-r1)(x-r2)





first you get


y = a (x-2) (x-7)





now the pt





-18 = a (5-2) (5-7)


-18 = a (3)(-2)


a = 3





final answer y = 3(x-2)(x-7)


you can expand the rest.
Reply:x^2-9x+14=y


Since the roots (answers) are 2 and 7, you can say y=(x-2)(x-7). Finish the multiplication, and get x2 -2x -7x +14 or x^2-9x+14.
Reply:Easy:





y = a(x - 2)(x - 7)





and (5, -18) satisfies this equation, so -18 = a(5 - 2)(5 - 7)


⇒ a = 3





So y = 3(x - 2)(x - 7) = 3x² - 27x + 42

phlox