How do you graph #f(x)= 5/3(x-3)^2 - 6#?

1 Answer
Aug 10, 2015

Calculate the vertex and the #x#- and #y#-intercepts and then sketch the graph.

Explanation:

#f(x)=5/3(x-3)^2-6#

Step 1. Your equation is in vertex form.

#f(x) = a(x-h)^2 +k#.

We see that #a=5/3#, #h=3#, and #k=-6#.

Step 2. Find the vertex.

The vertex is at (#h,k#) or (#3,-6#).

Step 3. Find the #y#-intercept.

Set #x=0# and solve for #y#.

#f(0)=5/3(x-3)^2-6 = 5/3(0-3)^2-6 = 5/3(-3)^2-6 = 5/3(9)-6=9#

The #y#-intercept is at (#0,9#).

Step 4. Find the #x#-intercept(s).

Set #f(x)=0# and solve for #x#.

#0=5/3(x-3)^2-6#

#5/3(x-3)^2 = 6#

#(x-3)^2= 6×3/5 =18/5#

#x-3=±sqrt(18/5)= =±sqrt((9×2×5)/(5×5))=±3/5sqrt10#

#x=3±3/5sqrt10#

#x=3+3/5sqrt10≈4.9# and #x=3-3/5sqrt10≈1.1#

Step 5. Draw your axes and plot the four points.

Graph1

Step 6. Draw a smooth parabola that passes through the four points.

Graph2

And you have your graph.