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#?

Question

Kennedy Adams · Accepted Answer

see 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]}

Genesis Allen · Answer

undefined To find the axis of symmetry of a quadratic function in the form $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex, the axis of symmetry is the vertical line passing through the vertex, given by $x = h$. In this case, the axis of symmetry is $x = 3$.

The graph of the function $y = 2(x - 3)^2 + 2$ is a parabola that opens upwards with its vertex at $(3, 2)$.

To find the maximum or minimum value of the function, you can observe that since the leading coefficient $a$ is positive, the parabola opens upwards, and the vertex represents the minimum value. So, the minimum value of the function is $y = 2$, which occurs at the vertex $(3, 2)$.