precalc question, please explain?

Question

find a function whose graph is a parabola with vertex (3 , 4) and that passes through the point (1, -8)

Answer 1

y = ax^2 + bx + c

x = (-b)/(2a)
3 = (-b)/(2a)
6a = -b
b = -6a

4 = a(3)^2 - 6a(3) + c
4 = 9a - 18a + c
4 = -9a + c
9a = c - 4
a = (1/9)(c - 4)

-8 = a(1)^2 - 6a(1) + c
-8 = a - 6a + c
-8 = -5a + c
5a = c + 8
a = (1/5)(c + 8)

(1/5)(c + 8) = (1/9)(c - 4)
9(c + 8) = 5(c - 4)
9c + 72 = 5c - 20
4c = -92
c = -23

a = (1/5)(c + 8)
a = (1/5)(-23 + 8)
a = (1/5)(-15)
a = -3

b = -6a
b = -6(-3)
b = 18

ANS : y = -3x^2 + 18x - 23

for a graph of my proof, go to http://www.calculator.com/calcs/GCalc.html

type it in like this -(3x^2) + 18x - 23

Answer 2

Knowing that one point is the vertex gives you a big hint. The general form of a parabola is:

y = a*x^2 + b*x + c

First, you can move your coordinate system to the vertes so (3, 4) becomes (0, 0) and (1, -8) becomes (-2, -12). I just subtracted 3 from each x value and 4 from each y. I will call the new coordinates u and v so they dont get confused with x and y. The relations between them are: u = x -3, v = y - 4

Now at the origin: 0 = a*0 + b*0 + c so: c=0

Since the vertex is at the origin, the parabola is symmetric about the v-axis meaning that the v value for any u must equal the v value for -u.:

a*u^2 + b*u = a*(-u)^2 + b*(-u) = a*u^2 - b*u

This can only be true if b=0.

The equation now is:

v = a*u^2

Use the other point (-2, -12):

-12 = a*(-2)^2 so: a = -3

So the equation is: v = -3 * u^2

Substitute the realtions for x and y in:

y - 4 = -3 * (x - 3)^2

y = -3 * (x^2 - 6*x + 9) + 4 = -3*x^2 + 18*x - 27 + 4

y = -3*x^2 + 18*x - 23

Of course check it on the two original points to see if it is right:

4 = -3*3^2 + 18*3 - 23 = 4

-8 = -3*1^2 + 18*1 -23 = -8

Both check out.