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

A) The function in vertex form is f(x) = (x – 4)2 – 11.
B) The vertex of the function is (–8, 5).
C) The axis of symmetry is x = 5.
D)The y-intercept of the function is (0, 5).
E) The function crosses the x-axis twice.

User Bmillare
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1 Answer

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Answer:

A, D, E are true

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.

User Michael Barrowman
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