Can anyone help me solve a math problem involving quadratic equations?

Question

Find the formula of a quadratic equation whose vertex is (-3,9) and passes through the point (1,41).

(The awkward wording is due to my shoddy spanish-to-english translation.) I have no idea how to do this, but I did manage to work out another coordinate: (-7,41), although I'm not sure if this is correct. Thanks for all help!

Answer 1

Actually, your wording is quite good - much better than many of the other questions I see.

As with most equations asking you to find a polynomial passing through a given set of points, you substitue the x and y-coordinates into the general polynomial equation Ax²+Bx+C=y to generate a system of linear equations that you can solve for A, B, and C. Two of the equations are obvious:

9A-3B+C=9
A+B+C=41

(these come from substituting (x=-3, y=9) and (x=1, y=41), respectively). Of course there's a problem - there are only two equations and three unknowns. The third equation comes from the fact that (-3, 9) is a vertex, so the derivative of the equation at -3 must be zero. The derivative of Ax²+Bx+C is 2Ax+B, and substituting x=-3 and setting it equal to zero gives you:

-6A+B=0

This gives you a system of three equations, that you may then solve for A, B, and C. You get:

A=2
B=12
C=27

So your quadratic equation is y=2x²+12x+27.

Answer 2

The general formula of a quadratic equation can be expressed as
Y = A(X + B)^2 + C

Since the turning point ( -3 , 9 ) is below the point ( 1 , 41 ), this quadratic function is " U " shaped.
In other words, the value of A is positive.

Since the minimum point is ( -3 , 9 ), B = 3 and C = 9

The reason :
(X + B)^2, a square function, is greater or equal to zero.
Since ( -3 , 9 ) is the minimum point, the lowest value for Y is 9.
That is why C has to be 9.

Also, when Y is 9, (X + B)^2 must be zero.
Since the x-coordinate is -3 when the y-coordinate is 9, B must be 3.

The desired quadratic equation is starting to take shape
Y = A(X + 3)^2 + 9

To find the value of A, we substitute the coordinates (1, 41) into the equation Y = A(X + 3)^2 + 9,
41 = A(1 + 3)^2 + 9
41 = 16 A + 9
32 = 16 A

A = 2

Hence the equation of the quadratic equation is Y = 2(X + 3)^2 + 9

You can also expand it to yield
Y = 2 X^2 + 12X + 27

Yes, (-7 , 41) is a point on this quadratic function. You will find that the coordinates of this point satisfy the equation.

I hope this helps

Answer 3

y= a(x + 3 ) (x + 3 )+9
y = a(x - p )(x - p ) + k
then plog the point to find the value of a
(1,41)
41 = a(1 +3 )(1 + 3 )+9
a = 2 and the equation will be
y =2(x + 3).(x+3) +9