+130511 We have that
y = ax^2 + bx + c we know that the point (0, 10) is on the graph
And we have another point on the graph (120, 10)...so we have
10 = a(120)^2 + b(120) + 10 → 0 = 14400a + 120b
We also know that the parbola is symmetric....so the max occurs at x = (120 + 0) / 2 = 60
And the max height = 100....so the point (60, 100) is also on the graph...so we have
100 = a(60)^2 +b(60) + 10 → 90 = 3600a + 60b
So using....
14400a + 120b = 0 and
3600a + 60b = 90
Multiply the second equation by -2 and add it to the first....this gives
7200a = -180
a = -1/40
So, solving for b we have ...... 3600(-1/40) + 60b = 90 → -90 +60b = 90 → 60b = 180 → b = 3
So....our function is
y =(-1/40)x^2 + 3x + 10
You can see the graph here.....https://www.desmos.com/calculator/zqusuapft0
I forgot that you wanted this in vertex form......we have......
-40y = x^2 - 120x - 400 →
-40y + 400 = x^2 - 120x .... complete the square.....
-40y + 400 + 3600 = x^2 - 120x + 3600 → (factor both sides)
-40(y - 100) = (x - 60)^2
+118723 +130511 a) x^2-2x-3=0 ; x^2-2x-3> 0
Factoring x^2-2x-3=0 we have that (x -3)(x +1) =0....and setting each factor to 0, we have that x = 3 and x = -1
For the second part, the polynomial will be greater than 0 either between x= -1 and x = 3 or in the two intervals on either side of this one. If we let x = 0 in the polynomial, note that it makes its value = -3. And this is less than 0. And since 0 falls in the "middle" interval above, the two "outside" intervals will make the inequality true. Thus, x < -1 and x > 3 are the two intervals that "work."
b) x^2-2lxl-3=0 ; x^2-2lxl-3> 0
This one is a little trickier. Let's start by rearranging x^2-2lxl-3=0 ......add 2lxl to both sides....this gives us
x^2 - 3 = 2lxl divide both sides by 2....so we have
(x^2 - 3 )/ 2 = l x l
We have two equations here....either ... (x^2 - 3 )/ 2 = x or - (x^2 - 3 )/ 2 = x
Working with the first one, multiply both sides by 2...we have...
x^2 - 3 = 2x rearranging gives us
x^2 -2x - 3 = 0 and the solution to this is the same as in (a) above. Notice something, however.....x =3 is OK in the original equation, but x = -1 doesn't "work" In effect, we have an extraneous solution.
Now, for the second part, we have - (x^2 - 3 )/ 2 = x ...multiply both sides by -2...this gives us
x^2 - 3 = -2x add-2x to both sides
x^2 +2x - 3 = 0 factoring gives (x +3) (x - 1) = 0 so either x = -3 or x = 1. Note that x = 1 is also extraneous, but x = -3 is OK.
Here's a graph of the equation....https://www.desmos.com/calculator/7rptum7sk3
Notice that it = 0 only at x = -3 and x = 3. This graph will also help us solve the second part of (b). Notice that the function is > 0 on the intervals x< -3 and x > 3. The odd thing about this graph is that it almost looks like a parabola except for the "kink" at the "vertex." This is due to the absolute value in the function.
c) lx^2-2x-3l=3
To save some time, I'm just going to present the graphs of y = lx^2-2x-3l and y = 3. Their intersection points will give us a solution.
Here are the graphs......https://www.desmos.com/calculator/9gcr9w4tay
Notice that the intersection points occur at about x = -1.646, x = 0, x = 2 and x = 3.646. We could have proved this algebraically, but I'm running out of gas, at this point !!!
