Find a Parabola given 2 points

Question

Find a formula for the parabola through the points (4,5) and (7,6)

Answer 1

(4,5), (7,6)

for a standard parabola, it could be vertical or horizontal

even an infinite number of both types

as well as an infinite number more of non-standard parabolas at an angle other than zero or 90 with the x axis

plug the two points into

y =x^2+bx +c, for the most simple standard parabola

5=16+4b +c

6=49+7b +c

subtract 1st from 2nd

1 =33 +3b

b = -32/3

c = 5-16+4(32/3) = -11+128/3 = 95/3

the parabola's equation is:

y= x^2 -32x/3 + 95/3

y= a(x-h)^2 +k is the standard vertical equation for a parabola. if a>0, it's upward opening, <0 downward opening, where (h,k) is the vertex and extreme point

let a=1 and solve for h and k, after plugging in the two points to find

the most simple parabola in vertex form

but those two points could also be on a horizontal parabola, right or left opening, as well as a vertical down ward opening parabola

for a horizontal plug the points into

x =y^2 +by + c

4= 25 +5b +c

6 =49+7b +c

subtract to get

2 = 24+2b

b =-22/2 =-11

c =-21 -5(-11)= 34

x =y^2 -11y+34 is the standard horizontal parabola

endless more standard parabolas by arbitrarily picking an "a" value other than 1 then solving for b and c

by plugging in the two points again into the general equations:

y= ax^2 +bx +c

or

x =ay^2 +by +c

Answer 2

5 = a(4²) + 4b + c

6 = a(7²) + 7b + c

Use techniques from Algebra 2 to solve the system.