The point P(x, y) is on the terminal ray of angle theta. If theta is between pi and 3pi/2 radians and csc theta equals -5/2 what
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You need 3 points to graph a quadratic function.
It is advisable that one of these points be the vertex, because the line of symmetry passes through it
Next, you need to find two other points that lie on the curve, replacing an x-value into the function and calculating the y-value. It is easier if these two points are at the same distance from the line of symmetry.
For example, if the quadratic function is y = (x - 4)² + 3, the vertex is located at (4, 3). Replacing with x = 3 into the equation, we get:
y = (3 - 4)² + 3
y = 1 + 3
y = 4
Replacing with x = 5 into the equation, we get:
y = (5 - 4)² + 3
y = 1 + 3
y = 4
Now, we need to connect the points (3, 4), (4, 3), and (5, 4) with a U shape, as follows:
The distance from point Y to the y-axis is 4 units and the distance from point Z to the y-axis is 3 units, then the lelgth of the segment YZ is 4+3=7 units.
If <span>a scale factor is 3, then the length of Y'Z' will be 7·3=21 units.
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P.S. In the added picture you can see trapezoid ABCD that was dilated by a scale factor 3 about the origin. This may help to understand that all linear values after dilation become multiplied by scale factor.
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If <span>a scale factor is 3, then the length of Y'Z' will be 7·3=21 units.
</span>
P.S. In the added picture you can see trapezoid ABCD that was dilated by a scale factor 3 about the origin. This may help to understand that all linear values after dilation become multiplied by scale factor.
<span> </span>