Produce a third degree polynomial that has exactly the roots -3 and 5 with y-intercept 1350.
I understand I could start with (x+3)(x-5)^2, but i don't know how to get the y-intercept 1350 with the same conditions. Please help. thank you.
you are only given two roots, so let the equation be
y = (x+3)(x-5)(x-a)
but you know that (0,1350) lies on it
1350 = (3)(-5)(-a)
1350 = 15a
a = 90
so equation is
y = (x+3)(x-5)(x-90)
expand if you have to, it is more useful in factored form.
btw, your equation of y = ....(x-5)^2 would have a double root of 5, that is, it would "touch" the x-axis at x = 5
To find a third-degree polynomial with the given conditions, we can start with the factored form you provided: (x+3)(x-5)^2.
To find the equation that passes through the y-intercept 1350, we need to determine the scale factor that will make the equation go through the point (0, 1350).
Let's substitute x=0 into the equation and solve for the scale factor (let's call it a):
(0+3)(0-5)^2 * a = 1350
3 * 25 * a = 1350
75a = 1350
a = 1350/75
a = 18
Now, we multiply the original equation by the scale factor a:
(18)(x+3)(x-5)^2
This will give us a third-degree polynomial that satisfies the conditions: has the roots -3 and 5, and passes through the y-intercept 1350.
The final equation is: 18(x+3)(x-5)^2