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 \(\displaystyle x=2y^{2}+5y-8\).