The equation of a horizontal parabola, in vertex form, is

$\displaystyle x = a(y-k)^{2}+h$,

where $\displaystyle (h,k)$ is the vertex. Set $\displaystyle h = -3,k=-6$:

$\displaystyle x = a[y-(-6)]^{2}+ (-3)$

$\displaystyle x = a(y+6)^{2}-3$

To find $\displaystyle a$, use the known $\displaystyle y$-intercept, setting $\displaystyle x =0, y=4$:

$\displaystyle 0 = a(4+6)^{2}-3$

$\displaystyle 0 = a\cdot 10^{2}-3$

$\displaystyle 0 =100 a-3$

$\displaystyle 100a= 3$

$\displaystyle a = \frac{3}{100}$

The equation, in vertex form, is $\displaystyle x = \frac{3}{100}(y+6)^{2}-3$; in standard form:

$\displaystyle x = \frac{3}{100}(y^{2}+12+36)-3$

$\displaystyle x = \frac{3}{100} y^{2}+\frac{9}{25} y+\frac{27}{25} - \frac{75}{25}$

$\displaystyle x = \frac{3}{100} y^{2}+\frac{9}{25} y-\frac{48}{25}$

A horizontal parabola which passes through  $\displaystyle (0,2)$ and $\displaystyle (0,6)$ has as its equation

$\displaystyle x = a(y-2)(y-6)$.

To find $\displaystyle a$, substitute the coordinates of the third point, setting $\displaystyle x=3,y=-2$:

$\displaystyle 3= a(-2-2)(-2-6)$

$\displaystyle 3= a(-4)(-8)$

$\displaystyle 3= 32a$

$\displaystyle a = \frac{3}{32}$

The equation is therefore $\displaystyle x = \frac{3}{32}(y-2)(y-6)$, which is, in standard form:

$\displaystyle x = \frac{3}{32}(y^{2}-8y+12)$

$\displaystyle x = \frac{3}{32} y^{2}- \frac{3}{4} y+ \frac{9}{8}$

A vertical parabola which passes through $\displaystyle (2,0)$ and $\displaystyle (6,0)$ has as its equation

$\displaystyle y = a(x-2)(x-6)$

To find $\displaystyle a$, substitute the coordinates of the third point, setting $\displaystyle x=3,y=-2$:

$\displaystyle -2 = a(3-2)(3-6)$

$\displaystyle -2 = a \cdot 1 \cdot (-3)$

$\displaystyle -2 = -3a$

$\displaystyle a = \frac{-2}{-3}= \frac{2}{3}$

The equation is $\displaystyle y = \frac{2}{3}(x-2)(x-6)$; expand to put it in standard form:

$\displaystyle y = \frac{2}{3}(x^{2}-8x+12)$

$\displaystyle y = \frac{2}{3}x^{2}-\frac{16}{3}x+8$

The equation of a horizontal parabola, in standard form, is

$\displaystyle x = ay^{2}+ by + c$

for some real $\displaystyle a,b,c$. 

$\displaystyle c$ is the $\displaystyle x$-coordinate of the $\displaystyle x$-intercept, so $\displaystyle c=6$, and the equation is

$\displaystyle x = ay^{2}+ by + 6$

Set $\displaystyle x=0, y=2$:

$\displaystyle 0 = a\cdot 2^{2}+ b \cdot 2 + 6$

$\displaystyle 0 = 4a+2b + 6$

However, no other information is given, so the values of $\displaystyle a$ and $\displaystyle b$ cannot be determined for certain. The correct response is that insufficient information is given.

The standard form of the equation of a vertical parabola is

$\displaystyle y=ax^{2}+bx+c$

If the values of $\displaystyle x$ and $\displaystyle y$ from each ordered pair are substituted in succession, three equations in three variables are formed:

$\displaystyle a\cdot (-2)^{2}+b(-2)+c = 23$

$\displaystyle 4a-2b+c = 23$

$\displaystyle a\cdot 1^{2}+b \cdot 1+c = 5$

$\displaystyle a+b+c = 5$

$\displaystyle a\cdot 3^{2}+b \cdot 3+c =13$

$\displaystyle 9a+3b+c = 13$

The system

$\displaystyle 4a-2b+c = 23$

$\displaystyle a+b+c = 5$

$\displaystyle 9a+3b+c = 13$

can be solved through the elimination method.

First, multiply the second equation by $\displaystyle -1$ and add to the third:

$\displaystyle 9a+3b+c =13$

$\displaystyle \underline{ -a-b-c \; \; \; =- 5 }$

$\displaystyle 8a+2b$         $\displaystyle =8$

Next, multiply the second equation by $\displaystyle -1$ and add to the first:

$\displaystyle 4a-2b+c = 23$

$\displaystyle \underline{ -a-b-c \; =- 5 }$

$\displaystyle 3a-3b$          $\displaystyle =18$

Which can be divided by 3 on both sides to yield

$\displaystyle a-b = 6$

Now solve the two-by-two system

$\displaystyle 8a+2b = 8$

$\displaystyle a-b = 6$

by substitution:

$\displaystyle a = b+ 6$

$\displaystyle 8 (b+6)+2b = 8$

$\displaystyle 8b+48+2b = 8$

$\displaystyle 10b+48 = 8$

$\displaystyle 10b= -40$

$\displaystyle b= -4$

Back-solve:

$\displaystyle a-(-4) = 6$

$\displaystyle a+4= 6$

$\displaystyle a =2$

Back-solve again:

$\displaystyle 2+(-4)+c = 5$

$\displaystyle -2+c=5$

$\displaystyle c=7$

The equation of the parabola is therefore 

$\displaystyle y = 2x^{2}-4x+7$.

First, find the $\displaystyle x$-intercepts—the points of intersection with the $\displaystyle x$-axis—of the lines by substituting 0 for $\displaystyle y$ in both equations.

$\displaystyle 2x+4y = 8$

$\displaystyle 2x+4 (0) = 8$

$\displaystyle 2x = 8$

$\displaystyle x=4$

$\displaystyle (4,0)$ is the $\displaystyle x$-intercept of this line. 

$\displaystyle 4x-2y = 12$

$\displaystyle 4x-2 (0)= 12$

$\displaystyle 4x= 12$

$\displaystyle x= 3$

$\displaystyle (3,0)$ is the $\displaystyle x$-intercept of this line. 

The parabola has $\displaystyle x$-intercepts at $\displaystyle (3,0)$ and $\displaystyle (4,0)$, so its equation can be expressed as 

$\displaystyle y = a(x-3)(x-4)$

for some real $\displaystyle a$. To find it, substitute using the coordinates of the third point, setting $\displaystyle x= 6, y = -4$:

$\displaystyle -4 = a(6-3)(6-4)$

$\displaystyle -4 = a \cdot 3 \cdot 2$

$\displaystyle -4 = 6 a$

$\displaystyle a = \frac{-4}{6} = -\frac{2}{3}$.

The equation is $\displaystyle y = -\frac{2}{3}(x-3)(x-4)$, which, in standard form, can be rewritten as:

$\displaystyle y = -\frac{2}{3}(x^{2}-7x+12)$

$\displaystyle y = -\frac{2}{3}x^{2}+\frac{14}{3}x-8$

The standard form of the equation of a horizontal parabola is

$\displaystyle x=ay^{2}+by+ c$

If the values of $\displaystyle x$ and $\displaystyle y$ from each ordered pair are substituted in succession, three equations in three variables are formed:

$\displaystyle a \cdot (-1)^{2}+b \cdot (-1)+ c = -11$

$\displaystyle a -b + c = -11$

$\displaystyle a \cdot (1)^{2}+b \cdot 1+ c = -1$

$\displaystyle a+b + c = -1$

$\displaystyle a \cdot (2)^{2}+b \cdot 2+ c = 10$

$\displaystyle 4a+2b+ c = 10$

The three-by-three linear system

$\displaystyle a -b + c = -11$

$\displaystyle a+b + c = -1$

$\displaystyle 4a+2b+ c = 10$

can be solved by way of the elimination method. 

$\displaystyle b$ can be found first, by multiplying the first equation by $\displaystyle -1$ and add it to the second:

$\displaystyle a+b + c = -1$

$\displaystyle \underline{-a+b - c = 11}$

        $\displaystyle 2b$         $\displaystyle =10$

$\displaystyle 2b \div 2 = 10 \div 2$

$\displaystyle b = 5$

Substitute 5 for $\displaystyle b$ in the last two equations to form a two-by-two linear system:

$\displaystyle a+5+ c = -1$

$\displaystyle a+5+ c - 5= -1 - 5$

$\displaystyle a + c = -6$

$\displaystyle 4a+2 \cdot 5+ c = 10$

$\displaystyle 4a+10+ c = 10$

$\displaystyle 4a+10+ c- 10 = 10 - 10$

$\displaystyle 4a +c = 0$

The system 

$\displaystyle a + c = -6$

$\displaystyle 4a +c = 0$

can be solved by way of the substitution method;

$\displaystyle 4a +c = 0$

$\displaystyle 4a +c -4a = 0 -4a$

$\displaystyle c = -4a$

$\displaystyle a + (-4a) = -6$

$\displaystyle -3a = -6$

$\displaystyle -3a\div (-3) = -6 \div (-3)$

$\displaystyle a = 2$

Substitute 2 for $\displaystyle a$ in the top equation:

$\displaystyle 2 + c = -6$

$\displaystyle 2 + c - 2= -6 - 2$

$\displaystyle c= -8$

The equation is $\displaystyle x=2y^{2}+5y-8$.

A circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ has equation

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

The circle has center $\displaystyle (2,4)$ and radius 5, so substitute:

$\displaystyle (x-2)^{2}+ (y-4)^{2}= 5^{2}$

$\displaystyle (x-2)^{2}+ (y-4)^{2}= 25$

A circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ has equation

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\displaystyle \left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\displaystyle \left ( \frac{-3+9}{2}, \frac{7+5}{2} \right )$

$\displaystyle (3, 6)$

Therefore, $\displaystyle h = 3$ and $\displaystyle k = 6$

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using $\displaystyle (3, 6)$ and $\displaystyle (9, 5)$. We will concern ourcelves with finding the square of the radius $\displaystyle r^{2}$:

$\displaystyle r^{2}= \left ( x_{2}-x_{1}\right )^{2} + \left ( y_{2}-y_{1}\right )^{2}$

$\displaystyle r^{2}= \left (9-3\right )^{2} + \left ( 5-6\right )^{2}$

$\displaystyle r^{2}=6^{2} + \left (-1\right )^{2}$

$\displaystyle r^{2}=36+ 1= 37$

Substitute: 

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

$\displaystyle (x-3)^{2}+ (y-6)^{2}=37$

A circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ has equation

$\displaystyle (x-h)^{2}+ (y-k)^{2}= r^{2}$

The circle has center $\displaystyle (-3, 2)$ and radius 4, so substitute:

$\displaystyle (x-(-3))^{2}+ (y-2)^{2}= 4^{2}$

$\displaystyle (x+3)^{2}+ (y-2)^{2}= 16$

$\displaystyle x^{2}+6x + 9+ y ^{2}-4y+4= 16$

$\displaystyle x^{2}+6x + 9+ y ^{2}-4y+4- 16= 16-16$

$\displaystyle x^{2}+6x + y ^{2}-4y-3=0$