Expert: Paul Klarreich - 10/30/2006

Question
Hi,

As I said before thank very much for giving the solulion for x intercept and y-intercept for equation y=-x^2+4x-5. But I failed to get the graph as per your instruction, can you please assist to me a graph? I will appreciate your reply.

Answer
Questioner:  Hassan
Category:  Advanced Math
Private:  no
 
Subject:  Question on graph
Question:  Hi,

As I said before thank very much for giving the solulion for x intercept and y-intercept for equation y=-x^2+4x-5.

But I failed to get the graph as per your instruction, can you please assist to me a graph? I will appreciate your reply.
............................................
Hi, Hassan,

I don't recall the question or instructions, but if you don't have a graphing calculator or a computer=based graphing program, (I think you can find some of these on the Web) then you just have to use the old graphing instruments -- a pencil and graph paper.  Then you do a little analysis:

For y = -x^2 + 4x - 5

A. The graph is a parabola which will open downward, because the leading coefficient, a = -1, is negative.

B. The x-intercepts are the solutions to the equation  -x^2 + 4x - 5.  Alas, the solutions are imaginary, so the graph will NOT cross the x-axis.

C. The  vertex has  x = -b/2a = 2.

NOW you set up a table:

x  |  y
---+----
  |
  |
  |
  |
  |
  |
  |
  |

Under the x-column you write 7 (seems a nice number) values:
three of which are on the left side of  x = 2,
three are on the right side of x = 2,
and one of them IS x = 2.

Under the y-column you write the computed y-values for each of those.

On your graph paper you plot the 7 points.

Then you connect them with a nice smooth curve that goes a little BEYOND the points.

That's your graph.