Question

The functions f(x) = -(x+4) +2 and g(x) =(x-2)^2 -2 have been rewritten using the completing the square method. Apply your knowledge of functions in vertex form to determine if the vertex for each function is a minimum or a maximum and explain your reasoning

Answer 1

Answer:

$g(x) = (x-2)^2 -2$

If we compare this function with the vertex form we see that:

$a = 1, h =2, k =-2$

So then the vertex would be $(h,k)=(2,-2)$

And since the value for a is positive we know that the parabola open upward. We don't have a maximum defined since open upwards and the minimum point correspond to the vertex on this case (2,-2).

$f(x)= -(x+4)^2 +2$

If we compare this function with the vertex form we see that:

$a=-1, h =-4, k = 2$

And since the value for a is negative we know that the parabola open downward. We don't have a minimum defined since open downwards and the maximum point correspond to the vertex on this case (-4,2).

Step-by-step explanation:

We need to remember that the standard form for a parabola is given by the following equation:

$y = ax^2 + bx +c$

And the vertex form is given by this formula:

$y = a(x-h)^2 +k$

And we want to find the vertex and if we have a maximum or minimum for each function. Let's begin with g(x)

$g(x) = (x-2)^2 -2$

If we compare this function with the vertex form we see that:

$a = 1, h =2, k =-2$

So then the vertex would be $(h,k)=(2,-2)$

And since the value for a is positive we know that the parabola open upward. We don't have a maximum defined since open upwards and the minimum point correspond to the vertex on this case (2,-2).

For the function f(x) we assume that we have the following equation:

$f(x)= -(x+4)^2 +2$

If we compare this function with the vertex form we see that:

$a=-1, h =-4, k = 2$

And since the value for a is negative we know that the parabola open downward. We don't have a minimum defined since open downwards and the maximum point correspond to the vertex on this case (-4,2).