Hello, mibamars!
Here's the first one . . .
For the parabola whose range is $\displaystyle y \leq 3$,
whose xcoordinate of the turning point is 4,
and whose yintercept is $\displaystyle \frac{7}{3}$
Find:
(1) the ycoordinate of the turning point.
(2) the equation of the parabola
(3) the coordinates of the xintercepts.
The general equation of a parabola is: .$\displaystyle y \;=\;ax^2+bx + c$
Make a sketch . . . Code:

(4,3) 
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        +     
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Since $\displaystyle \left(0,\,\frac{7}{3}\right)$ is on the parabola: .$\displaystyle \frac{7}{3} \:=\:a\!\cdot\!0^2 + b\!\cdot\!0 + c\quad\Rightarrow\quad \boxed{c \:=\:\frac{7}{3}}$
The equation (so far) is: .$\displaystyle y \;=\;ax^2 + bx  \frac{7}{3}$
Since the range (values of y) is < 3,
. . and the xcoordinate of the turning point is 4,
. . clearly the vertex is (4,3).
$\displaystyle {\color{red}\text{(1) The }y\text{coordinate of the turning point is }3.}$
The xcoordinate of the vertex is given by: .$\displaystyle \frac{b}{2a}$
. . So we have: .$\displaystyle \frac{b}{2a} \:=\:4\quad\Rightarrow\quad 8a  b \:=\:0$ . [1]
Since (4,3) is on the parabola: .$\displaystyle 3 \:=\:a(\text{}4)^2 + b(\text{}4)  \frac{7}{3}$
. . which simplifies to: .$\displaystyle 16a  4b \:=\:\frac{16}{3}\quad\Rightarrow\quad 4a  b \:=\:\frac{4}{3}$ . [2]
Subtract [2] from [1]: .$\displaystyle 4a \:=\:\frac{4}{3}\quad\Rightarrow\quad\boxed{ a \:=\:\frac{1}{3}}$
Substitute into [1]: .$\displaystyle 8\left(\text{}\frac{1}{3}\right)  b \:=\:0\quad\Rightarrow\quad\boxed{ b \:=\:\frac{8}{3}}$
$\displaystyle {\color{red}\text{(2) The equation of the parabola is: }\:y \;=\;\frac{1}{3}x^2  \frac{8}{3}x  \frac{7}{3}}$
For the xintercepts: .$\displaystyle \frac{1}{3}x^2  \frac{8}{3}x  \frac{7}{3} \;=\;0$
Multiply by 3: .$\displaystyle x^2 + 8x + 7 \:=\:0$
. . Factor:.$\displaystyle (x+7)(x+1) \:=\:0\quad\Rightarrow\quad x \:=\:7,1$
$\displaystyle {\color{red}\text{(3) The }x\text{intercepts are: }\:(7,0), (1,0)}$