# What are the intercepts of 2x - 5x^2 = -3y +12?

Jul 13, 2018

$y$ intercept $\left(0 , 4\right)$

no $x$-intercepts

#### Explanation:

Given: $2 x - 5 {x}^{2} = - 3 y + 12$

Put the equation in $y = A {x}^{2} + B y + C$

Add $3 y$ to both sides of the equation: $\text{ } 2 x - 5 {x}^{2} + 3 y = 12$

Subtract $2 x$ from both sides: $\text{ } - 5 {x}^{2} + 3 y = - 2 x + 12$

Add $5 {x}^{2}$ from both sides: $\text{ } 3 y = 5 {x}^{2} - 2 x + 12$

Divide both sides by $3 : \text{ } y = \frac{5}{3} {x}^{2} - \frac{2}{3} x + 4$

Find the $y$ intercept by setting $x = 0 : \text{ } y = 4$

Find the $x$ intercepts by setting $y = 0$ and using the quadratic formula:
$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$

$x = \frac{\frac{2}{3} \pm \sqrt{\frac{4}{9} - \frac{4}{1} \cdot \left(\frac{5}{3}\right) \cdot \frac{4}{1}}}{\frac{2}{1} \cdot \frac{5}{3}} = \frac{\frac{2}{3} \pm \sqrt{\frac{4}{9} - \frac{80}{3}}}{\frac{10}{3}}$

$x = \frac{\frac{2}{3} \pm \sqrt{- \frac{236}{9}}}{\frac{10}{3}}$

No Real solutions since $x$ is imaginary (negative square root).