Please graph the function.  `f(x) = (x+3)^2 + 2`

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`f(x) = (x+3)^2 + 2`

The given function is an equation of the parabola in vertex form:

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

where (h,k) is the vertex and the value of a indicates the direction of the parabola.

If a is positive the parabola opens upward. And opens downward if a is negative.

To clarify the value of h and a, let's re-write the function as:

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

So the vertex of the parabola is (-3, 2) and it opens upward.

Since it is upward, the vertex is the minimum point of the function.

To graph, we still need additional two points.

To determine the other points, assign values of x that is less than h and also greater than h. Then, solve for y.

So at `(-oo, -3]` , let `x=-6` .

`y=(-6+3)^2+2 = (-3)^2+2=9+2=11`

And at `[-3,+oo)` , let `x = 0` .

`y=(0+3)^2 + 2 = 3^2+2=9+2=11`

Then plot the points (-6,11), (-3,2) and (0,11). Connect them, following the shape of the parabola.

Hence, the graph of `f(x) = (x+3)^2+2` is:

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