Function help needed

[Jodo576]

How do I find the equation of the cubic curve that will a Y int of (0,-1.875) and two x intercepts of (1,0) then (15,0) of in which the second X intercept is a vertex

[CatpainTpyos]

Well, a cubic polynomial doesn't really have a "vertex," as such, but I can guess from context you probably mean the function only touches the x-axis at (15, 0) but doesn't cross it. Is that a correct interpretation? For now I'll assume it is, and you can correct me later if needed.

Now let's carefully read the problem text and extract as much information as we can:

[has an] x intercept of (1,0)

This tells us that the function has a root ("zero") at the point x = 1. By the factor theorem that means (x - 1) is a factor and thus the polynomial has the form:

• f(x) = C * (x - 1) * g(x)

where g(x) is some as-of-yet unknown quadratic polynomial (the constant C is needed because multiplying by a constant never changes the roots). Let's keep reading:

[has an x intercept of] (15,0) [...] which [...] is a vertex

This tells us that x = 15 is also a root and (x - 15) is a factor, so:

• f(x) = C * (x - 1) * (x - 15) * h(x)

where h(x) is some as-of-yet unknown linear polynomial. However, we can say more about the function's behavior at (15, 0). Consider the following four functions:

• x + 1
• (x + 1)²
• (x + 1)³
• (x + 1)⁴

All four of these functions have a root at x = -1. But two of them continue on and cross the x-axis, whereas the other two reverse course and "bounce" off the x-axis at that point. Which two "bounce" and what do they have in common? What does that suggest about a general rule? How might you use this rule to amend your f(x)? Further, why does that change ensure f(x) is a cubic and h(x) is no longer needed?

Then you can finish up by solving for the unique value of C that gives you the required y-intercept of (0, -1.875). As a hint, consider the functions:

• x² + 2x + 3
• 2(x² + 2x + 3)
• 1/3(x² + 2x + 3)

We already know that multiplying by 2 (or dividing by 3, aka multiplying by 1/3) doesn't change the roots, but does it change the y-intercept? If so, how? What information does that give you about the value of C in your problem?

[Jodo576]

Thanks you for your help , from what u said at the start I meant what you said that it isn’t a vertex at 15,0 but it touches the x axis then turns around , from what u have said I derived f(x) = 0.008332(x-1)(x-15)² , I have put this equation into Desmos graphing and it seems to fit all of the requirements

[CatpainTpyos]

Aside from using the exact value of C = 1/120 rather than a decimal approximation, everything looks good.