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

1 Answer
Jul 13, 2018

Answer:

#y# intercept #(0, 4)#

no #x#-intercepts

Explanation:

Given: #2x - 5x^2 = -3y + 12#

Put the equation in #y = Ax^2 + By + C#

Add #3y# to both sides of the equation: #" "2x - 5x^2 + 3y = 12#

Subtract #2x# from both sides: #" "- 5x^2 + 3y = -2x + 12#

Add #5x^2# from both sides: #" "3y = 5x^2 -2x + 12#

Divide both sides by #3: " "y = 5/3x^2 - 2/3x + 4#

Find the #y# intercept by setting #x = 0: " "y = 4#

Find the #x# intercepts by setting #y = 0# and using the quadratic formula:
#x = (-B+- sqrt(B^2 - 4AC))/(2A)#

#x = (2/3 +- sqrt(4/9 - 4/1 * (5/3)* 4/1))/(2/1 * 5/3) = (2/3 +- sqrt(4/9 - 80/3))/(10/3)#

#x = (2/3 +- sqrt(-236/9))/(10/3)#

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