How do you write #f(x) = -4x^2 - 16x + 3# in vertex form?
1 Answer
Jun 18, 2017
Explanation:
#"for the standard form of a parabola " y=ax^2+bx+c#
#"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)#
#y=-4x^2-16x+3" is in standard form"#
#"with " a=-4,b=-16,c=3#
#rArrx_(color(red)"vertex")=-(-16)/(-8)=-2#
#"substitute into f(x) for y-coordinate"#
#rArry_(color(red)"vertex")=-4(-2)^2-16(-2)+3=19#
#rArrcolor(magenta)"vertex "= (-2,19)#
#"the equation of a parabola in "color(blue)"vertex form"# is.
#• y=a(x-h)^2+k# where ( h , k ) are the coordinates of the vertex and a is a constant.
#"here " (h,k)=(-2,19)" and "a=-4#
#rArry=-4(x+2)^2+19larrcolor(red)" in vertex form"#