Question

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

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

Answer 1

May I help?

Answer 2

yes please!!

Answer 3

That is with my pleasure!
What do you think?
Any ideas?

Answer 4

the function crosses twice, the axis system is x=5, the y inercept of the function is (0,5)

Answer 5

i think thats what it is im not sure though

Answer 6

Did you try to plot it?

Answer 7

@trip21

Do you follow?

Answer 8

Thank you for the medal @skullpatrol!

Answer 9

almost everything right, but the axis of symmetry is not x = 5
Let's run through each statement one by one.
f(x) = x2 – 8x + 5

The function in vertex form is f(x) = (x – 4)2 – 11.
Complete the square on f(x) to show if this is true or not (or expand this one here)
(x-4)^2 - 11 = x^2 - 8x + 16 - 11 = x^2 - 8x + 5 = f(x). so the statement is true.

The vertex of the function is (–8, 5).
For general form, vertex is given by x = -b/2a, and its y-value given by f(-b/2a).
x = -(-8)/2(1) = 8/2 = 4, thus the vertex is at 4, not -8.
However when f(x) has been completed as a square (vertex form): (x-h)^2 + k, then the vertex is given by (h, k), so reading off the vertex form we get (4, -11)

The axis of symmetry is x = 5.
See above, axis of symmetry is at x = -b/2a = 4, not 5.

The y-intercept of the function is (0, 5).
correctamundo. In general form, the constant term is always the y-intercept (check by subbing in x = 0)

The function crosses the x-axis twice.
There are a few ways to test this. One way is to find the discriminant.
as b^2 - 4ac = (-8)^2 - 4(1)(5) = 64 - 20 = 44 > 0.
If the discriminant > 0, then we have two distinct solutions, thus the graph has two x-intercepts which we do see in this case. Discriminant = 0 means the graph contacts the x axis just once at the vertex exactly, while discriminant < 0 means the graph never contacts the x axis which we describe as a positive or negative definite.

Another way is to use the vertex form (or simply see the value of the maximum or minimum)
(x-4)^2 - 11 is a squared number minus 11. Squared real numbers can be at smallest zero. Thus the smallest value we have is 0 -11 = -11. The graph is also concave up (the coefficient of x^2 is positive). Our minimum (vertex) is below the x axis and climbs up, thus we do expect two pts of contact with the x axis.

The sketch is shown below