How do you find the vertex and intercepts for #y=2(x-3)^2 +4#?
1 Answer
Apr 10, 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=2(x-3)^2+4" is in this form"#
#"with " a=2, h=3" and " k=4#
#rArrcolor(magenta)"vertex "=(3,4)#
#"since " a>0" then min. turning point "uuu#
#color(blue)"Intercepts"#
#• " let x = 0, in equation, for y-intercept"#
#• " let y = 0, in equation, for x-intercepts"#
#x=0toy=2(-3)^2+4=22larrcolor(red)" y-intercept"#
#y=0to2(x-3)^2+4=0#
#rArr(x-3)^2=-2# This has no real solutions hence f(x) has no x-intercepts.
graph{2(x-3)^2+4 [-31.56, 31.67, -15.8, 15.8]}
