GIVING BRAINILEST-
1 answer:
Answer:
To construct a dilation, we need the center of dilation, which is the point from which we make the image smaller or larger. We also need a scale factor, which is the number that determines how large or small we are making the new image. If the scale factor is greater than 1, we have an enlargement, because the new image will be larger than the original image. If the scale factor is less than 1, we have a reduction because the new image will be smaller than the original image.
Enlargement
Step-1
Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.
Step-2
First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.
Step-3
Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.
Step-4
Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.
Step-5
Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.
We now repeat the process for each of the other vertices of the triangle.
Step-6
These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.
Step-7
Finally, we use the straightedge to connect all of the new vertices.
Now,See how triangle P
′
Q
′
R
′
is twice the size of the original triangle.
Reduction
Step-1
Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of
2
1
Step-2
Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.
Step-3
The green lines are the new lines that we have drawn.
Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.
Step-4
Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.
Step-5
Now, we will take the straightedge and connect the points where the curves overlap.
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Answer:
Step-by-step explanation:
The domain of all polynomials is all real numbers. To find the range, let's solve that quadratic for its vertex. We will do this by completing the square. To begin, set the quadratic equal to 0 and then move the -10 over by addition. The first rule is that the leading coefficient has to be a 1; ours is a 2 so we factor it out. That gives us:
The second rule is to take half the linear term, square it, and add it to both sides. Our linear term is 2 (from the -2x). Half of 2 is 1, and 1 squared is 1. So we add 1 into the parenthesis on the left. BUT we cannot ignore the 2 sitting out front of the parenthesis. It is a multiplier. That means that we didn't just add in a 1, we added in a 2 * 1 = 2. So we add 2 to the right as well, giving us now:
The reason we complete the square (other than as a means of factoring) is to get a quadratic into vertex form. Completing the square gives us a perfect square binomial on the left.
and on the right we will just add 10 and 2:
Now we move the 12 back over by subtracting and set the quadratic back to equal y:
From this vertex form we can see that the vertex of the parabola sits at (1,-12). This tells us that the absolute lowest point of the parabola (since it is positive it opens upwards) is -12. Therefore, the range is R={y|y ≥ -12}