How do you find the vertex and the intercepts for #y = x^2 + 6x − 16#?

1 Answer
Sep 16, 2017

#"see explanation"#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where (h , k ) are the coordinates of the vertex and a is a multiplier.

#"to obtain this form use "color(blue)"completing the square"#

#• " ensure coefficient of "x^2" term is 1"#

#• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2+6x#

#rArry=x^2+2(3)xcolor(red)(+9)color(red)(-9)-16#

#color(white)(rArry)=(x+3)^3-25#

#rArrcolor(magenta)"vertex "=(-3,-25)#

#color(blue)"Intercepts"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercepts"#

#x=0toy=-16larrcolor(red)" y-intercept"#

#y=0to(x+3)^2-25=0#

#rArr(x+3)^2=25#

#color(blue)"take the square root of both sides"#

#rArrx+3=+-5larr" note plus or minus"#

#rArrx=+-5-3#

#rArrx=2" or "x=-8larrcolor(red)" x-intercepts"#