How do you graph #y = -.3270 + 13.74x – 4.7x^2#?

1 Answer
Jun 7, 2015

You can graph this using a graphing utility...or by knowing a few tricks about quadratic equations!

First, use the quadratic formula to find the roots (or "zeros") of the equation:

#x=(-color(red)b+-sqrt(color(red)b^2-4color(blue)acolor(green)c))/(2color(blue)a)#

I've colored your original equation to show which coefficients plug into the quadratic formula here:

#y=-color(green)0.3270+color(red)13.74x-color(blue)4.7x^2#

Plug those coefficients into the quadratic formula for the following:

#x=(-color(red)13.74+-sqrt((-color(red)13.74)^2-4(-color(blue)4.7)(-color(green)0.3270)))/(2(-color(blue)4.7))#

When you simplify, you get two roots: #x~=0.024# and #x~=2.90#. Mark these two points on your graph.

Next, observe that the sign of the #color(blue)a# coefficient is negative, so the direction of the parabola is curved down. Assuming you cannot use calculus to find the maximum, you can instead plug in values between the two roots until you find where it changes direction. The following table shows that the graph turns between #x=1.4620# and #x=1.7496#:

Matlab R2015a, Home Edition

So the max must be somewhere between #9.3# and #9.7#. Plot a max at about #(1.5,9.5)#, then draw your parabola through the two roots and the max. It should look like this:

graph{y=-0.3270+13.74x-4.7x^2 [-2,5,-10,20]}