Question

Part 1: Given y = x2 + 4x – 12 determine the following. Use complete sentences and show all work to receive full credit. Does the graph open up or down? How do you know?----The graph open upward Explain whether the graph has a maximum or minimum point.------The graph has minimum point Find the vertex and x-intercepts of the graph. The graph has vertex at P(-2,-16) and x- intercept@ x=2 and x=-6 Part 2: Create your own unique quadratic equation in the form y = ax2 + bx + c that opens the same direction and shares one of the x-intercepts of the graph of y = x2 + 4x − 12. Determine the following

Answer 1

@tcarroll010

Answer 2

Sorry wrong problem this is the true problem
Part 1:
Solve each of the quadratic equations below and describe what the solution(s) represent to the graph of each. Show your work to receive full credit.

0 = x2 + 5x + 6
0 = x2 + 4x + 4
Part 2:
Using complete sentences, answer the following questions about the two quadratic equations above.
Do the two quadratic equations have anything in common? If so, what?
What makes y = x2 + 4x + 4 different from y = x2 + 5x + 6?

Answer 3

Each of these quadratic equations can be solved by factoring. Each of these equations is in the form:\[ax ^{2} + bx + c = 0\]where "a" is "1". This is rather advantageous because solving the equations has become greatly simplified. For each, look for 2 numbers that when multiplied, give you "c" but when added, give you "b". This might seem hard at first, but it will become easy if you first list all pairs of numbers that are the factors of "c". Do that now, write that out here, and then we'll proceed to the next step.

Answer 4

I'm still confused

Answer 5

np. Start with the first equation. "c" is "6". List out pairs of numbers that when multiplied give you 6.

Answer 6

3,2
-2,-3
1,6
-1,-6

Answer 7

That's great! Now among those 4 pairs, list out all pairs that add to 5

Answer 8

1,6

Answer 9

no I mean 3,2

Answer 10

Great again! The reason we did this is because we use these 2 numbers, call them "a" and "b", to solve the equation.

(x + a)(x + b) = 0 -> (x + 2)(x + 3) = 0

because (x + 2)(x + 3) = x^2 + 5x + 6 = 0

We can more easily find the solution in the factored form because any number times "0" is "0". So our solution to equation #1 is

x = -2 or x = -3.

Answer 11

Oh ok now I get it

Answer 12

So why did we even bother? Part of the initial questions ask why this is important to graphing. We have "y = 0" at those 2 x's, so we can graph 2 points so far:

(-2, 0) and (-3, 0)

Answer 13

Those 2 points are called the "zeros" of the graph or equation (where y = 0) and it is at those values of "x".

Answer 14

So for 0 = x2 + 4x + 4

4=

2,2
1,4
-2,-2
-1,-4

2+2 = 4

(2,0) and (0,2)

Answer 15

A good amount of your last post is correct. Up to and including 2 and 2. That's good work! But the rest goes like this:

(x + 2)(x + 2) = 0 -> (x + 2)^2 = 0

This is zero only at x = -2. That's the point (-2, 0)

Answer 16

So, what these 2 equations have in common is that they both open upward and are the same shape: Neither one is wider or narrower than the other. They are only in different places on the graph. They are shifted in relation to each other. What is different is that the first graph intersects the x-axis (where y=0) in 2 places, whereas the second just touches the x-axis in one place. If you like, I can get you a really good graph of both but it will take about 2 minutes. Would you like that?

Answer 17

yes please

Answer 18

Notice how the shape and width are the same for both

Answer 19

oh ok it makes more sense now

Answer 20

thanks for all your help

Answer 21

uw! Good luck to you in all of your studies and thx for the recognition! @yellowsun

Answer 22

your welcome!