How do you find the axis of symmetry, graph and find the maximum or minimum value of the function #y-2=2(x-3)^2#?
1 Answer
see explanation.
Explanation:
#" Express " y-2=2(x-3)^2" in the form"#
#rArry=2(x-3)^2+2# 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+2" is in this form"#
#"here " h=3" and " k=2#
#rArr" vertex "=(3,2)# To determine min/max consider the value of a
#• a>0rArr" minimum " uuu#
#• a<0rArr" maximum " nnn#
#"here " a=2rArr" minimum"# The axis of symmetry passes through the vertex and is vertical with equation
#color(blue)(x=3)# The minimum value at the vertex is y = 2
#color(blue)"Intercepts"#
#x=0toy=2(-3)^2+2=20larrcolor(red)" y-intercept"#
#y=0to2(x-3)^2=-2#
#rArr(x-3)^2=-1# which has no real solutions and therefore graph does not cross the x-axis.
graph{(y-2x^2+12x-20)(y-1000x+3000)=0 [-40, 40, -20, 20]}
