How do you find the vertex and intercepts for y=x^2+4x+1y=x2+4x+1?

1 Answer
Oct 10, 2017

(-2,-3),x=-2+-sqrt3(2,3),x=2±3

Explanation:

"the equation of a parabola in "color(blue)"vertex form"the equation of a parabola in 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 the coefficient of "x^2" term is 1"

• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2+4x

y=x^2+4x+4larr" coefficient of "x^2" term is 1"

y=x^2+2(2)xcolor(red)(+4)color(red)(-4)+1

color(white)(y)=(x+2)^2-3

rArr"vertex "=(-2,-3)

color(blue)"Intercepts"

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

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

x=0toy=4-3=1larrcolor(red)" y-intercept"

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

rArr(x+2)^2=3

rArrx+2=+-sqrt3larr" note plus or minus"

rArrx=-2+-sqrt3larrcolor(blue)"exact values"

x~~ -3.73,x~~ -0.27larrcolor(red)" x-intercepts"