Question

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(– 2, – 16), and x intercepts at x = -6 and x = 2. (Do not include the negative sign in your answer.)

y = x2 + 4x − ___

This problem has been frustrating me, pretty difficult.

Answer 1

If I could know how to give someone a medal, I would.
If they could explain to me how I would work this out and the formula.

Answer 2

Well, if (-2, -16) is the vertex, then it is on our graph, right? That means that, it is a solution to the equation.
So, if we plug in the point, it will give us a true statement.

Answer 3

So, could I use a graphing software for this? Or what should I plug in to the equation or point?

Answer 4

y = x^2 + 4x − ___ <-- let's call the blank space some variable, 'c'. We will solve for c.
y = x^2 + 4x + c So, (x, y) = (-2, -16); x=-2 and y=-16. We'll only have the 'c' to solve for after substituting these two other values. :)

Answer 5

So use the quadratic formula: x = -b + or - b^2-4ac/2a or should I use this formula x=-b/2a

Answer 6

a = 1 b = 4 and c is blank

Answer 7

Well, the first only helps for intercepts, which 'could' work but might take more work; and the second is for the vertex, which we were given.

Here, I'm just using the original equation and the given point to solve for what the blank is going to be. Any point on the graph of that equation makes the equation true when we plug it in.

Answer 8

y = x^2 + 4x − __ blank will be called c
y = x^2 + 4x − c Let (x, y) = (-2, -16);
-16 = (-2)^2 + 4(-2) - c <-- can you see what I did here with the substitution?

Answer 9

We can evaluate those numerical expressions now:
-16 = 4 - 8 - c
-16 = -4 - c <-- it should be more clear that this is a single variable equation here.

Answer 10

I know but what would I do to solve it

Answer 11

By regurally using factoring trinomials?

Answer 12

OH but, if any equatin on the graph of thaat equation makes it true how is that possible when it is graded by a computer and cannot have multiple answeres.

Answer 13

I do not follow what you are asking about...

Answer 14

AccessDenied is correct and has provided you with the solution and showing you step by step on how to arrive to the correct solution

Answer 15

not sure where the mistake is but I used a regression on my graphing calculator and c should be a negative value, I just fixed the sign on c to make sure c ends up negative....

Answer 16

The original question shows a subtraction of the blank space, so I arranged for c to be subtracted accordingly. That way, c was exactly what went into the blank space. :)

Answer 17

clever, I missed that then you are 100% correct.......