Hello, ty2391!
Another approach . . .
Determine the equation of the parabola with xintercepts 6 and 2, and a yintercept of 4.
The general form of the parabola is; .$\displaystyle y \:=\:ax^2 + bx + c$
Since the yintercept is 4, the point $\displaystyle (0,4)$ satisfies the equation.
. . We have: .$\displaystyle 4 \:=\:a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad c = 4$
The equation (so far) is: .$\displaystyle y \:=\:ax^2 + bx + 4$
Here is a sketch of the parabola . . . Code:

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6 2  2

We know that the axis of the parabola is halfway between the xintecepts.
. . That is, the vertex lies on the vertical line $\displaystyle x = 2$
We also know that the vertex is at $\displaystyle x \:=\:\frac{\text{}b}{2a}$
. . Then: .$\displaystyle \frac{\text{}b}{2a}\:=\:2\quad\Rightarrow\quad b \,= \,4a$
So the equation is: .$\displaystyle y \:=\:ax^2 + 4ax + 4$
Since $\displaystyle (2,0)$ satisfies the equation, we have:
. . $\displaystyle 0 \:=\:a\cdot2^2 + 4a\cdot2 + 4\quad\Rightarrow\quad 12a \,=\,\text{}4\quad\Rightarrow\quad a = \text{}\frac{1}{3}$
Therefore, the equation is: .$\displaystyle \boxed{y \;= \;\frac{1}{3}x^2  \frac{4}{3}x + 4}$