Sketch y=|(x-1)(x-3)| and y=|x^2-4x| .
I don't know how to sketch the graphs :(
in your previous post I showed you how to graph the absolute value of linear function.
The same concept is true for the above quadratics
for y=|(x-1)(x-3)|
You will have two parabolas, one is
y = (x-1)(x-3) the other is y = -(x-1)(x-3)
one will open up , the other will open down
They both have the same x-intercepts of 1 and 3
So sketch both of them, then erase the part of both of them that lies below the x-axis.
Do the same for the second equation.
They will have similar shapes, except the second has x-intercepts of 2 and -2
i still don't really understand. How can i sketch y=(x-1)(x-3) and y=-(x-1)(x-3) ? what is the coordinate?
To sketch the graphs of y = |(x-1)(x-3)| and y = |x^2-4x|, we can follow a step-by-step process. Here's how you can do it:
1. Plot the x-intercepts:
- For the equation y = |(x-1)(x-3)|, set the expression inside the absolute value equal to zero:
(x-1)(x-3) = 0
This yields x = 1 and x = 3 as x-intercepts.
- For the equation y = |x^2-4x|, set the expression inside the absolute value equal to zero:
x^2-4x = 0
Factoring out x, we have x(x-4) = 0, which gives x = 0 and x = 4 as the x-intercepts.
2. Determine the sign of the expression within the absolute value:
- For y = |(x-1)(x-3)|, we know that both factors, (x-1) and (x-3), are non-negative when x ≥ 3 or x ≤ 1.
Hence, the expression within the absolute value will be positive or zero in these intervals, which results in |(x-1)(x-3)| = (x-1)(x-3).
In the interval 1 < x < 3, both factors are negative, and the expression within the absolute value becomes positive, resulting in |(x-1)(x-3)| = -(x-1)(x-3).
- For y = |x^2-4x|, we consider (x^2-4x) as a single factor. The quadratic expression is non-negative when its vertex (at -b/2a) is either on or to the right of the x-axis. In this case, the vertex is x = 2, which means x^2-4x ≥ 0 when x ≤ 0 or x ≥ 4.
Therefore, the expression within the absolute value will be positive or zero in these intervals, resulting in |x^2-4x| = x^2-4x.
In the interval 0 < x < 4, the quadratic expression is negative, and the expression within the absolute value becomes positive, leading to |x^2-4x| = -(x^2-4x).
3. Sketch the graphs:
- For y = |(x-1)(x-3)|:
• Starting from the left of the x-axis, when x ≤ 1, the function is positive or zero, taking the form y = (x-1)(x-3). Plot these points on the graph.
• When 1 < x < 3, the function is negative, so reflect the portion between x = 1 and x = 3 to the positive side of the y-axis.
• Finally, for x ≥ 3, the function is positive or zero, taking the form y = (x-1)(x-3). Plot these points on the graph.
- For y = |x^2-4x|:
• Starting from the left of the x-axis, when x ≤ 0, the function is positive or zero, taking the form y = x^2-4x. Plot these points on the graph.
• When 0 < x < 4, the function is negative, so reflect the portion between x = 0 and x = 4 to the positive side of the y-axis.
• Finally, for x ≥ 4, the function is positive or zero, taking the form y = x^2-4x. Plot these points on the graph.
Please note that this explanation assumes you have a basic understanding of plotting points on a graph.