y = 2x+3 + 4 cuts the y axis when x = 0
so y = 23 + 4 = 8 + 4 = 12.
Therefore, the point is (0, 12).
The domain of y = 2x is [0, +infinity].
please think about what you just said
When you plug in x = 0, you get y = 8. When you plug in x = 8, you get y = 0. So the graph of y = x - 8 (which is clearly linear, a straight line, because it only has x to the first power) is the graph of the straight line that goes through the points (0,8) and (8,0).
4
It has an absolute minimum at the point (2,3). It has no maximum but the ends of the graph both approach infinity.
no
The domain of y = 2x is [0, +infinity].
please think about what you just said
ex = x3 This has two solutions: x = 4.5364... and x = 1.85718... Plot the graph of each and you can see the intersections.
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?
It's almost like the parabola but y goes up more fast than y = x2
It depends on whether the value of the power.
When you plug in x = 0, you get y = 8. When you plug in x = 8, you get y = 0. So the graph of y = x - 8 (which is clearly linear, a straight line, because it only has x to the first power) is the graph of the straight line that goes through the points (0,8) and (8,0).
4
It has an absolute minimum at the point (2,3). It has no maximum but the ends of the graph both approach infinity.
Select a set of values of x for which you wish to plot the graph - a subset of the domain. For each value x, calculate y = negative x to the second power = -x2. Plot the set of ordered points (x, y) = (x, -x2) on a coordinate plane and jion them together with a smooth curve.
1.5848931924611135 to the power of five equals 10.