This is the long way:
x2 + 16 = 0
x1= -1/2 + squareroot of -15,75
x2 = -1/2 - sqareroot of - 15,75
Note: squareroot of 16,25 is roughly ~ 4,03112575
Well, so much for the theory, but this question does not have any solution, since you can't draw the squareroot of a negative number like - 15,75.
And the short way:
x2 + 16 = 0
x2 = -16
x = squareroot of - 16, which as I declared above, is not possible.
Woops, I misunderstood your question; heres the answer to your question:
f(x) = x2 + 16
derivate
f'(x) = 2x
f'(0) = 2(0) Put in x = 0
f'(0) = 0 When x = 0, the curve turns and you have a minimi point.
f(0) = x2 + 16 Put in x = 0 in the first equation to get y.
f(0) = 02 + 16
f(0) = 16 y = 16
minimum point: x = 0, y = 16 --> (0;16)
5 7
between A and B
2048
3
Yes
A graph that has 1 parabolla that has a minimum and 1 positive line.
It has an absolute minimum at the point (2,3). It has no maximum but the ends of the graph both approach infinity.
The minimum value of the parabola is at the point (-1/3, -4/3)
There is no maximum but te minimum is 15.
mid point of xy
5 7
The point (4, 5) is.
The vertex has a minimum value of (-4, -11)
y = x2 + 3 Since the x term is missing, the x-coordinate of the vertex is 0. If x = 0, then y = 3. Thus, (0, 3) is the vertex, the minimum point of the parabola.
you have to find the key with the plus and equals symbol. Then push the shift key and the plus/equals key. The "cross" should show up. Plus (+) equals (=). The plus should be above the equals.
When x = -2 then y = 4 which is the common point of intersection of the two equations.
the midpoint of AB.