change the y value to -y, and bring the negative over the equal sign. example. y=2x^2 reflected on the x-axis looks like y=(2x^2)/-1 which is equal to y=-(2x^2)
First you go over how many spaces then you go up how many spaces
The plus values become negative and the negative values become positive although their numerical values remain the same
The x value remains the same, but the y value changes sign
something which decreases over time e.g. speed of a car when the brakes are applied. Time on the x-axis and speed on the y-axis of the graph
The bit with the negative x-axis goes to the positive x-axis.
no
Reflecting a point over the x-axis involves changing the sign of the y-coordinate while keeping the x-coordinate the same. If a point is already located over the x-axis, its y-coordinate is positive. When reflecting this point over the x-axis, the positive y-coordinate becomes negative, resulting in the point being located below the x-axis.
by looking and controling it
reflect across the x-axis and then reflect again over the x-axis
You change the value of y to -y. ex: (4,5) reflected over the x-axis is (4,-5)
change the y value to -y, and bring the negative over the equal sign. example. y=2x^2 reflected on the x-axis looks like y=(2x^2)/-1 which is equal to y=-(2x^2)
When reflecting a point over the x-axis, you are essentially changing the sign of the y-coordinate while keeping the x-coordinate the same. So, if the original point has coordinates (x, -y), reflecting it over the x-axis would result in the new coordinates being (x, y). This transformation is a fundamental concept in geometry and can be applied to various shapes and figures to create mirror images across the x-axis.
x
count the spaces on your graph until you reach the y axis then start over and count again till you count the same number that you it took you to reach the y axis... sounds kinda confusing.... but good luck !
I dont really know if this is right but i think to do this problem you have to take a point then rotate the paper counter clockwise around the origin then you have a new point which is called a prime. Then reflect it over the y axis on the graph.
x=-b/2a [negative B over 2A]