Question

f(x)=-x^2+8x+15

Answer 1

sometimes

Answer 2

so what is the question?

Answer 3

what are you supposed to do with this function?

Answer 4

yeah lol...

Answer 5

graph

Answer 6

parabola that opens down with vertex at (4,31)
http://www.wolframalpha.com/input/?i=-x^2%2B8x%2B15

Answer 7

To plot the graph, you need to know the x-intercepts and y intercept, and then take note of the - sign in front of the x^2

Answer 8

Here i how I would do it:
\[f(x)=-x^2+8x+15\]
set f(x) = 0 to find the points at which the graph crosses the x-axis. So
\[-x^2+8x+15 = 0\]
multiply through by -1
\[x^2-8x-15 = 0\]
\[(x-4)^2 - 31 = 0\]
\[x = 4 \pm \sqrt{31}\]
So these are the points at which the graph corsses the x-axis. To find the point where it crosses the y-axis, set x=0 in your original equation to get 15. Now because of the negative on the x^2, your graph will be an upside down parabola, going through \[(0, 15), (4 - \sqrt{31}, 0) and (4 + \sqrt{31}, 0)\]
To find the coordinates of the maximum (it is a maximum) of the graph, you take a look at the completed square method above. Since we multiplied through by -1, we need to multiply through by it again to get:
\[f(x) = 31 -(x-4)^2\]
Now this is maximal when x=4, because x=4 causes -(x-4)^2 to vanish. So the coordinates of the maximum are (4,y). To find the y, simply substitute x=4 into the equation f(x) to give y = 31. So it agrees with the mighty Satellite: (4,31) is the vertex.

Answer 9

sorry that took so long to write, this laptop is from 2997 and makes me wait about 5 minutes in between scrolling up and down and takes 2 seconds to register that i have typed...

Answer 10

note that any other value of x (even a negative x) will cause (x-4)^2 > 0 meaning that at those points x will not be maximal.