graph f(x)

Question

Graph    f(x)= -x²+3x-8

 

Please help...graphing these equations is my weakness...im lost

Answer 1

To graph you have to find the following points:

 

   X intercepts:

 

   - X^2 + 3X + 8 =0

 

  X = 3/+2 +√((9- 4( -1) (8))/2

 

      = 3/2 ±√(41) /2

 

   Vertex:

            ( 3/2, 41/4)

 

     Y intercept ( 0, 8 )

 

  Always, like in this case ( a<0) the parabola opens down ward, and vertex is at maximum.

So you have the vertex

 

  Axis of Symmetry:

 

  X = 3/2

 

 Locate the vertex and locate the X intercept and draw the parabola in such a way that intercepts the

 

    Y- axis @( 0, 8).

 

Answer 2

This is the graph of an upside down(because negative sign in front) parabola.

 

Since this is standard form of a quadratic equation (f(x)=ax² + bx+ c), you can start by  finding the "x" part of the vertex:

 

Use the formula x=-b/2a =-3/2(-1) = 3/2

Now, substitute this x-vaule into original equation to find y-value of vertex:

y= -(3/2)² +3(3/2) -8

  = -9/4 + 9/2 -8

  = -9/4 + 18/4 - 32/4 (just got common denominators here so could add/subt fractions)

  = -23/4

So, vertex is (3/2,-23/4).

 

Next, try to find x-intercepts by setting F(x)=0 (this is where the parabola crosses x-axis):

If you can't factor without using quadratic formula, then try picking values on either side of the x-value of the vertex:

 

x                y     

1                -6    (when substitute x=1 in original equation, you get y=-6)

3/2      -23/4    (vertex)

2                -6    

 

Now, plot these points and should have an upside down parabola.