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"