How is the graph of #y=8x^2-1# different from the graph of #y=8x^2#?

2 Answers

See below:

Explanation:

With the function #y=8x^2# (or really any function), we're taking a series of values of #x#, dropping them into the function, and getting a #y# value.

In this case, when #x=0, y=0#, when #x=1, y=8#, and when #x=-10, y=800#

So let's now look at #y=8x^2-1# - how is it different? For each value of #x# that we put into this function, the resulting value of #y# will be one less than for the other function.

This gives us when #x=0, y=-1#, when #x=1, y=7#, and when #x=-10, y=799#.

Graphically, they look like this (with the #y=8x^2# nested just above the #y=8x^2-1#):

graph{(y-8x^2)(y-8x^2+1)=0}

Jun 24, 2017

See below

Explanation:

A simple and short answer is that the first equation goes through #{0,-1}# as the #y#-intercept, but the second equation's #y#-intercept is the origin.

graph{y=8x^2-1 [-10, 10, -5, 5]}

graph{8x^2 [-10, 10, -5, 5]}