Question

Find, in the form y=ax²+bx+c, the equation of a quadratic whose graph has vertex (-3, 1) and passes through (1, -11)

Answer 1

Hi there, I can probably solve this one.

Answer 2

Do you know how to solve simultaneous systems of linear equations ?

Answer 3

In this case you have three unknowns -- a, b, and c -- and so need 3 equations describing the relationship among them.

The first two are easy. You are given two points on the parabola so just substitute in the x and y values accordingly, to get your first two equations:

9a - 3b + c = 1
a + b + c = -11

For the third equation you need calculus. You are given one additional piece of information, which is that the vertex is at -3, 1 -- the vertex is the point at which the first derivative is 0. So differentiate the original equation; you get

b + 2ax

At x=-3 you know that's 0, so that's your third equation:

b = 0

Now solve everything simultaneously, you get

a = -3/4
b = -9/2
c = -23/4

Answer 4

Write y = a(x+3)^2+1 Now calculate a by replacing x and y with (1,-11)
-11 = a(1+3)^2+1
-12 = a(16)
a = -3/4
Equation is y = -3/4(x+3)^2 + 1 Put it in the required form
y = -3/4(x^2+6x+9)+1
y = -3/4x^2 -9/2x -23/4

Answer 5

Vertex=(-3,1)
Point on Parabola= (1,-11)
From the quadratic equation we know that the x-coordinate of the vertex is
\[x=-\frac{b}{2a}\]
So,
\[(1)\;\;\;\;-3=-\frac{b}{2a} \space \space \space so \space \space \space -6a=-b \space \rightarrow a=\frac{b}{6}\]

We have three unknowns so we need three equations, plug in each point into quadratic form equation to get the other two, so

\[(2)\;\;\;\;-11=a(1)^{2}+b(1)+c\]
\[(3)\;\;\;\;1=a(-3)^{2}+b(-3)+c\]

Then substitute accordingly and solve for each one, although MERTSJ solution is a lot faster.