The function V(t)= 1,000(1.06)^2t models the value of an investment after t years.
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In the picture we can see the graph represents of the function f(x)=x²-2x+3 that is curve graph.
Given that,
We have to find the graph represents the function f(x)=x²-2x+3.
We know that,
The quadratic function is f(x)=x²-2x+3.
This is a vertical parabola open downward
The vertex is a maximum
We know,
A vertical parabola's equation in vertex form is equal to
y= a(x-h)²+k
where
a is a coefficient
(h,k) is the vertex
Here
a=1
(h,k)=(0,3) is vertex
The y-intercept is the point (0,3) is value of y when the value of x is equal to zero (is the same point that the vertex)
The x-intercepts are the points is values of x when the value of y is equal to zero
Therefore, In the picture we can see the graph represents of the function f(x)=x²-2x+3.
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