What are the intercepts of #3x - 5y^2 = 6#?

1 Answer
Apr 15, 2018

Answer:

**#x# intercept: #(2, 0)#

#y# intercept: NONE**

Explanation:

Before we find the x intercept, let's first make #x# by itself:
#3x - 5y^2 = 6#

Add #5y^2# to both sides of the equation:
#3x = 6 + 5y^2#

Divide both sides by #3#:
#x = (6+5y^2)/3#

#x = 2 + (5y^2)/3#

To find the #x# intercept, we plug in #0# for #y#, and solve for #x#:
#x = 2 + (5(0)^2)/3#

#x = 2 + 0/3#

#x = 2 + 0#

#x = 2#

So we know that the #x# intercept is #(2, 0)#.


Now let's make #y# by itself to find the #y# intercept:
#3x - 5y^2 = 6#

Subtract #3x# from both sides of the equation:
#-5y^2 = 6 - 3x#

Divide both sides by #-5#:
#y^2 = (6-3x)/-5#

Square root both sides:
#y = +-sqrt((6-3x)/-5)#

Now plug in #0# for #x#:
#y = +-sqrt((6-3(0))/-5#

#y = +-sqrt(-6/5)#

Since you can't square root a negative number, that means the solution is imaginary, meaning that there is no #y# intercept.

To check that our intercepts are correct, we can graph this:enter image source here

As you can see from the graph, it never touches the #y# axis, meaning that there is no value of #y# when #x# is zero. Also, you can see that the #x# intersect is in fact #(2, 0)#.

Hope this helps!