i need help finding quadratic functions please?

Question

find the quadratic function that has the indicated properties: A. vertex: (-2,5); contains point (0,9)
B. vertex: (4,-1); contains point (2,3)

Answer 1

if it contains the point 0, 9 and vertex is -2,5...then it aso contains the point -4,9
sub these x,y values into y = ax^2 +bx +c for 3 equations
so 9 = a(0)^2 +b(0) +c
and 5 =a(-2)^2 +b(-2) +c
and 9 = a(-4)^2 +b(-4) +c
from 1st equation know that c = 9
so sub that into the other 2 equations
5 = 4a +-2b +9
and 9=16a -4b +9
so from those -4=4a-2b and 0=16a-4b
solve this sy elimination method and multiply -2(-4=4a-2b) and get 8=-8a+4b and add to 0=16a -4b
8=-8a+4b
0=16a-4b
8=8a
a=1
so 0=16(1) -4b
-16=-4b
b=4 and from begining c = 9 so the equation for problem a is
y=x^2 +4x +9...sub in (x,y) from given points to check

for problem b
if vertex is 4,-1 and 2,3 is a point then 6,3 is also a point
get 3 equations
-1 =a(4)^2 +b(4) +c
3=a(2)^2 +b(2)+c
3=a(6)^2 +b(6)+c
so -1 =16a+4b+c and 3=4a+2b+c and 3=36a+6b+c
and solve for the variables

Answer 2

(A) Let the quadratic function be : y = ax^2 + bx + c

If the point (0,9) lies on it, then by substitution we get :
9 = a * 0^2 + b * 0 + c
Therefore, c = 9.
The equation is now : y = ax^2 + bx + 9

A vertex occurs when the slope = 0.
Differentiating y = ax^2 + bx + 9 gives slope = 2ax + b.
Letting 2ax + b = 0, gives x = -b/(2a).

This is the x value at the vertex, which we know is -2.
Therefore, -2 = -b/(2a), or b = 4a.

Substituting this b value into y = ax^2 + bx + 9 gives :
y = ax^2 + 4ax + 9
which is the vertex point equation.

Now substitute the point (-2,5) into this equation to give :
5 = a(-2)^2 + 4a(-2) + 9
5 = 4a - 8a + 9
-4a = -4
a = 1
Substituting into b = 4a, gives : b = 4*1 = 4

The equation is therefore : y = x^2 + 4x + 9
------------------------------------
(B) Let the quadratic function be : y = ax^2 + bx + c

Point (2,3) lies on it, so : 3 = a(2)^2 + b(2) + c
Therefore, c = 3 - 4a - 2b
Equation is now : y = ax^2 + bx + 3 - 4a - 2b

Vertex is at x = -b/(2a), i.e., 4 = -b/(2a), so b = -8a

Substituting into the previous equation
gives the vertex point equation as :
y = ax^2 + (-8a)x + 3 - 4a - 2(-8a)
y = ax^2 - 8ax + 3 - 4a + 16a
y = ax^2 - 8ax + 3 + 12a

Substituting the point (4, -1) into this gives :
-1 = a(4)^2 - 8a(4) + 3 + 12a
-1 = 16a - 32a + 3 + 12a
-4a = -4
a = 1
Substituting into b = -8a gives : b = -8

Now, c = 3 - 4a - 2b = 3 - 4(1) - 2(-8) = 15

The equation is therefore : y = x^2 - 8x + 15

Answer 3

Given vertex (-2,5)
and quad fn contains point (0, 9)
Let quadratic fn. be,
Y=a*X^2 + b*X + c
( X^2 means X*X or X squared)
Now, Y contains point (0,9)
Therefore, 9=a*0*0 + b*0 + c
=> c=9

The vertex of quad. function Y=a*X^2 + b*X + c is
( -b/2a , -(b^2 - 4ac)/4a )
Thus,
-b/2a = -2 and -(b^2 - 4ac)/4a = 5

therefore, b=4a => b^2=16a^2
Thus putting this in -(b^2 - 4ac)/4a = 5 we get
-(16a^2 - 4ac)/4a = 5
putting value of c=9 we get
16a^2 - 4a*9 = -5*4a
16a^2 - 36a = -20a
16a^2 - 16a=0
a^2 - a=0
a cannot be equal to zero because then the equation will not be quadratic
So, a=1
Earlier, -b/2a = -2
Thus, b = 4a=4

Thus, a=1, b=4 and c= 9.
Thus the quadratic function is
Y=X^2 + 4*X + 9

Similarly you can solve for any number of vertex and containing points.

Answer 4

The way you should approach these problems is to set up a quadratic equation in the form y-k = a (x-h)^2, substituting x and y from the point contained and substituting the vertex into h and k. Then find the value a.

for example, you have vertex (1,2) and it contains point (3,18)
set up something like this: 18-2 = a (3-1)^2
then you find that a = 4
put that into the equation y-k = a (x-h)^2, so that you have h and k subbed in, and the a value subbed in...
resulting in y-18 = 4 (x-3)^2

try doing it like that...