How do you find the vertex and intercepts for #y = 2(x – 5)^2 + 2#?
1 Answer
Apr 29, 2017
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-5)^2+2" is in this form"#
#"with " (h,k)=(5,2)larrcolor(red)" vertex"#
#color(blue)" to find intercepts"#
#• " let x=0, in equation, for y-intercept"#
#• " let y=0, in equation, for x-intercept"#
#x=0toy=2(-5)^2+2=52larrcolor(red)" y-intercept"#
#y=0to2(x-5)^2+2=0#
#rArr(x-5)^2=(-2)/2=-1#
#(x-5)^2=-1" has no real solutions thus no x-intercepts"#
graph{2(x-5)^2+2 [-10, 10, -5, 5]}
