PLZZ HELP! Given the function f(x)=2(x-1)^2+3

a)What is the vertex?
b)What is the axis of symmetry?

a) (1,3)

b) x=1

Step-by-step explanation:

The general form of a parabola in vertex form is f(x)=a(x-h)^2+k, where h is the x coordinate of the vertex, and k is the y coordinate, while a indicates how "squished" the graph is, or how quickly it grows in relation to the parent graph. In this case, h is shown to be 1, and k is 3, meaning that the vertex is at (1,3). The axis of symmetry , for a vertical parabola, just runs vertically through the vertex, meaning that in this case it is at x=1. Hope this helps!

(1,3) and y-axis

Step-by-step explanation:

The vertex is the value of x at maximum or minimum point

At maximum point

dy/ dx = 0;

dy/dx = 4(x-1)= 0

=>x= 1 and is the vertex

Y when x = 1; we substitute x=1 in the original equation.

f(x) =y =2(x-1)^2+3

y = 2(1-1)^2 +3

y = 0 + 3= 3

The vertex is (1,3)

2.

A symmetrical equation is given by

b^2 = 4ac

Looking at the equation above x has an order of 2 meaning a possibility of cutting the x axis on two values of x

Hence the axis of symmetry is the y-axis

Assuming this was the expression; the axis of rotation would have been the x-axis.

f(y)=2(y-1)^2+3

the solution is 12x-6y