How do you write #y-5 = -2(x+1)^2# in intercept form?

1 Answer
Jan 19, 2018

Answer:

#y=(x+1+1/2sqrt10)(x+1-1/2sqrt10)#

Explanation:

#"arrange into standard form "y=ax^2+bx+c ;a!=0#

#y-5=-2x^2-4x-2#

#rArry=-2x^2-4x+3larrcolor(blue)"in standard form"#

#"with "a=-2,b=-4" and "c=3#

#"find the roots (x-intercepts) by equating to zero"#

#rArr-2x^2-4x+3=0#

#"use the "color(blue)"quadratic formula"#

#x=(4+-sqrt(16+24))/-4=(4+-sqrt40)/-4=(4+-2sqrt10)/-4#

#rArrx=-1+-1/2sqrt10#

#y=(x-(-1-1/2sqrt10))(x-(-1+1/2sqrt10))#

#rArry=(x+1+1/2sqrt10)(x+1-1/2sqrt10)#