Point K is located at (7,4) in the coordinate plane. Its image is created by the following rule:

Mathematics

1 answer:

Fynjy0 [20]2 years ago

6 0

We are given coordinate of K point (7,4).

We need to find the new coordinate of K' point.

Given rule is K' = Ro 270°^(k).

That represents rotation of k 270 degree about the origin.

<em>The rule for rotation of 270 degree about the origin is as following:</em>

<em>(x, y) --------> (y, -x).</em>

Now, applying same rule on point (7,4).

(7,4) -----------> (4,-7).

<h3>Therefore, the coordinates of k' is (4,-7).</h3><h3>Correct option is A. (4,-7).</h3>

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Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.

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Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A. Do this by first setting the function equal to 0, then moving the 5 to the other side.

$x^2-8x=-5$

Now we can complete the square. Take half the linear term, square it, and add it to both sides. Our linear term is 8 (from the -8x). Half of 8 is 4, and 4 squared is 16. So we add 16 to both sides.

$(x^2-8x+16)=-5+16$

We will do the addition on the right, no big deal. On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex. We will rewrite with that perfect square on the left and the addition done on the right,

$(x-4)^2=11$

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

$(x-4)^2-11=y$

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value. This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation. In the case of this type of parabola, that "x = " will always be equal to the h value. So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x. When we do that, and then solve for y, we find that when x = 0, y = 5. So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x. I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice. Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0). That's why you need the quadratic formula to find these values.

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Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

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Answer:

$Vertex = (\frac{1}{2},\frac{7}{4})$