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](https://tex.z-dn.net/?f=x%5E2-8x%3D-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](https://tex.z-dn.net/?f=%28x%5E2-8x%2B16%29%3D-5%2B16)
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](https://tex.z-dn.net/?f=%28x-4%29%5E2%3D11)
Now we will move the 11 back over, which gives us the k coordinate of the vertex.
![(x-4)^2-11=y](https://tex.z-dn.net/?f=%28x-4%29%5E2-11%3Dy)
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.