a=3. Now put the values. 4= 3-b+c, -b+c=1. Vertex occurs when dy/dx=0.

This gives equation dy/dx= 6x+b. b=6 at x=-1, c= 1+b=c=7

I get the equation y=3x^2 +6x +7

When you have a quadratic (y = ax² + bx + c), the x-coordinate of the vertex is:

x = -b/(2a)

If you need help remembering this, it's essentially the quadratic formula, but without the ±√ part.

You know your x-coordinate is -1, and a = 3:

-1 = -b/(2*3)

-1 = -b/6

-6 = -b

b = 6

So your revised equation is:

y = 3x² + 6x + c

Plug in your point (-1,4) --> x = -1, y = 4 and solve for c:

4 = 3(-1)² + 6(-1) + c

4 = 3 - 6 + c

4 = -3 + c

4 + 3 = c

c = 7

Final equation:

y = 3x² + 6x + 7

Answer:

a = 3

b = 6

c = 7

Let's start with the vertex form of a quadratic:

y = a(x - h)² + k

Where the vertex is (h, k).  We are told this is (-1, 4) so we can now set that to be:

y = a(x + 1)² + 4

We can now expand this:

y = a(x² + 2x + 1) + 4

y = ax² + 2ax + a + 4

Now we can look at your equation:

y = 3x² + bx + c

We can now compare the coefficients of the like-terms between the two equations and set them equal:

a = 3 and 2a = b and a + 4 = c

We now have a system of three equaitions and three unknowns.  We know the value of "a" now so we can substitute and simplify to solve for b and c:

2a = b and a + 4 = c

2(3) = b and 3 + 4 = c

6 = b and 7 = c

So the values of a, b, and c are:

a = 3, b = 6, and c = 7