Find b and c so that y=4x^2+bx+c has vertex (-7,6)

Well, to find the vertex of a quadratic, we have a few options. we can either complete the square, or solve for the x value of the vertex using a nifty equation. I'm gonna go that route, but if your interested in the complete the square thing I can show you.

Anyway, the x value of a vertex equals

x = -b/2a

assuming

f (x) = ax^2 + bx + c.

so b is the coefficient in front of just x and a the coefficient in front of x^2 and so on. let's solve for x in your problem.

y = 4x^2 + bx + c

so, our x value has to be

x = -b/2 (4) right? so how do we solve for b, which is what we wanna know, from that? well, we KNOW the x value is supposed to be -7 in this case. so that means...

-7 = -b/2 (4) now just solve for b.

b = 56 if you need the work I can show you in comments.

Anyway, now we know b. how do we solve for c? well, let's fill in our old equation with our new info.

y = 4x^2 + 56x + c

what do we know? well, we know the vertex of this line is (-7,6). what does this tell us? well, the vertex is a point on our line, so that means I could plug in my point, right?

6 = 4 (-7)^2 + 56 (-7) + c

if (-7,6) is a point on my line, ^^^ that will be true. now all you gotta do is solve for c. if you need the work, I can show ya in the comments.

c = -196

hope this helps!

1.17+3.99a. you combine the terms with a to create 3.99 and the 1.17 stays because it has no like terms or a variable