# How do you find the vertex and intercepts for #y = 2(x + 4)^2 - 21#?

##### 1 Answer

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=2(x+4)^2-21" is in this form"#

#"with " h=-4" and " k=-21#

#rArrcolor(magenta)"vertex "=(-4,-21)#

#color(blue)"to find intercepts"#

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

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

#x=0toy=2(4)^2-21=32-21=11larrcolor(red)" y-intercept"#

#y=0to2(x+4)^2-21=0#

#rArr2(x+4)^2=21#

#rArr(x+4)^2=21/2#

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

#sqrt((x+4)^2)=color(red)(+-)sqrt(21/2)larr" note plus or minus"#

#rArrx+4=+-sqrt(21/2)#

#"subtract 4 from both sides"#

#xcancel(+4)cancel(-4)=+sqrt(21/2)-4#

#rArrx=-4+-sqrt(21/2)larrcolor(red)" x-intercepts"#

#"these are the exact values, which may be approximated to"#

#x=-7.24" and " x=-0.76" to 2 dec. places"#

graph{2(x+4)^2-21 [-46.3, 46.16, -23.07, 23.18]}