Question

Find 3 quadratic equations that go through (0,0) and (6,0).

I have found two. I cannot find a third.

1. f(x)=x(x-6)
2. f(x)= -x(x-6)

Answer 1

This one works too f(x)=x^2(x-6)

Answer 2

Thank you, but it needs to be a parabola. I appreciate your input though!

Answer 3

given 2 points, there are an infinite number of quadratics that wil fit

Answer 4

define a vertex; (a,b) to build the vertex format with

y = k(x-a)^2+b

now for any given points: (p,q) and (s,t) we can setup a system of equations

q = k(p-a)^2+b
t = k(s-a)^2+b

Answer 5

or do a matrix run

Answer 6

notice that your x-intercepts, thhat is, your endpoints are

Answer 7

these are roots arent they :/

Answer 8

given 2 roots: a,b

y = k(a-x)(b-x) satisfies all quadratics with roots a,b for any real k

Answer 9

so use the parabola "vertex form" and pick a point, as @amistre64 said, between those 2 given points for the vertex

once you get those coordinates for the vertex between those 2 points, plug in your ( 6, 0) values and solve for "a" \(\bf\large y = \color{red}{a}(x-\textit{x-coord here})^2+\textit{y-coord here}\)

Answer 10

say for example, let's pick... say between 0 ... 6 for "x" say.... (3, -5) down below
\(\bf y = \color{red}{a}(x-3)^2-5\qquad x = 6\qquad y = 0\qquad (6,0)\qquad thus\\ \quad \\
y = \color{red}{a}(x-3)^2-5\implies 0 = \color{red}{a}(6-3)^2-5\)

Answer 11

as and @amistre64 said..... over teh x=3 line, you can pick an infinite values for "y" to get infinite amount of vertices and thus infinite parabola functions

Answer 12

Bottom line f(x)=2x(x-6) will do

Answer 13

im not sure i like that bottom line ... its just to flat