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

Apr 15, 2018

**$x$ intercept: $\left(2 , 0\right)$

$y$ intercept: NONE**

#### Explanation:

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

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

Divide both sides by $3$:
$x = \frac{6 + 5 {y}^{2}}{3}$

$x = 2 + \frac{5 {y}^{2}}{3}$

To find the $x$ intercept, we plug in $0$ for $y$, and solve for $x$:
$x = 2 + \frac{5 {\left(0\right)}^{2}}{3}$

$x = 2 + \frac{0}{3}$

$x = 2 + 0$

$x = 2$

So we know that the $x$ intercept is $\left(2 , 0\right)$.

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

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

Divide both sides by $- 5$:
${y}^{2} = \frac{6 - 3 x}{-} 5$

Square root both sides:
$y = \pm \sqrt{\frac{6 - 3 x}{-} 5}$

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

$y = \pm \sqrt{- \frac{6}{5}}$

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: 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 $\left(2 , 0\right)$.

Hope this helps!