Line A (y = 3x – 3) is transformed into
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To know the range of the equation, you must first see the graph by plotting arbitrary points. Set values of x in any interval and substitute it to the equation. You would get a corresponding value of f(x) or y. When you plot the points, the graph would look like that in the picture.
The range of the graph refers to the coverage of y-values of the curve. From the picture, we could see that it covers most of the negative y-axis. To know at which value, let's find the maxima. There is no minima for this graph as you can see because it extends downwards infinitely at both sides.
To find the maxima, differentitate f(x) with respect to x and equate to 0
f'(x) = 0 = -20-2x-12x^3=0
You can find the roots easily by using a scientific calculator using the mode EQN. For this equation, there are 2 complex root and 1 real root. Thus, x = -1.14. Substitute this to the original equation:
y = 13 - 20(-1.14) - (-1.14)^2 -3(-1.14)^4
y = 29.433
Thus, the maxima is at point (-1.14,29.433). It can't be seen clearly in the graph because of the close intervals.
The range of the graph refers to the coverage of y-values of the curve. From the picture, we could see that it covers most of the negative y-axis. To know at which value, let's find the maxima. There is no minima for this graph as you can see because it extends downwards infinitely at both sides.
To find the maxima, differentitate f(x) with respect to x and equate to 0
f'(x) = 0 = -20-2x-12x^3=0
You can find the roots easily by using a scientific calculator using the mode EQN. For this equation, there are 2 complex root and 1 real root. Thus, x = -1.14. Substitute this to the original equation:
y = 13 - 20(-1.14) - (-1.14)^2 -3(-1.14)^4
y = 29.433
Thus, the maxima is at point (-1.14,29.433). It can't be seen clearly in the graph because of the close intervals.