How do you find the vertex and intercepts for #y=2(x-3)^2 +4#?

Question

Isaiah Anthony · Accepted Answer

see 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]}

Abigail Allen · Answer

undefined To find the vertex of the parabola represented by the equation $ y = 2(x - 3)^2 + 4 $, we use the formula for the vertex of a parabola in the form $ y = a(x - h)^2 + k $, where $ (h, k) $ is the vertex. In this equation, $ h = 3 $ and $ k = 4 $, so the vertex is at $ (3, 4) $.

To find the x-intercepts, we set $ y = 0 $ and solve for $ x $. 
$$
0 = 2(x - 3)^2 + 4 \implies 2(x - 3)^2 = -4 \implies (x - 3)^2 = -2
$$
This equation has no real solutions because a square cannot be negative.

To find the y-intercept, we set $ x = 0 $ and solve for $ y $.
$$
y = 2(0 - 3)^2 + 4 \implies y = 2(9) + 4 \implies y = 18 + 4 \implies y = 22
$$
So, the y-intercept is at $ (0, 22) $.