Write three decimals that are equivalent
1 answer:
You might be interested in
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.
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.
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,
Now we will move the 11 back over, which gives us the k coordinate of the vertex.
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.
1+tan^2(A) = sec^2(A) [Pythagorean Identities]
tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)
*see photo for complete solution*
tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)
*see photo for complete solution*