How do you find the intercept and vertex of y-4 = -(x-1)^2?

1 Answer
May 15, 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 constant.

y-4=-(x-1)^2

"add 4 to both sides"

ycancel(-4)cancel(+4)=-(x-1)^2+4

rArry=-(x-1)^2+4larrcolor(red)" in vertex form"

"with " h=1" and " k=4

rArrcolor(magenta)"vertex "=(1,4)

color(blue)"Intercepts"

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

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

x=0toy=-(0-1)^2+4=3larrcolor(red)" y- intercept"

y=0to-(x-1)^2+4=0

rArr-(x-1)^2=-4

"multiply both sides by - 1"

rArr(x-1)^2=4

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

sqrt((x-1)^2)=+-sqrt4larrcolor(red)" note plus or minus"

rArrx-1=+-2larr" add 1 to both sides"

rArrx=1+-2

rArrx=1-2=-1,x=1+2=3larrcolor(red)" x- intercepts"
graph{-(x-1)^2+4 [-10, 10, -5, 5]}