Math  /  Algebra

Question(9) Point of symmetry of the function f(x)=1x+3+2f(x)=\frac{1}{x+3}+2 is \qquad (a) (3,2)(3,-2) (b) (3,2)(-3,-2) (c) (3,2)(3,2) (d) (3,2)(-3,2)

Studdy Solution

STEP 1

What is this asking? Where's the center of this cool graph, the point it flips over perfectly? Watch out! Don't mix up the *x* and *y* values!
The signs can be tricky too.

STEP 2

1. Analyze the Base Function
2. Apply Transformations
3. Identify the Point of Symmetry

STEP 3

Let's start with our *base function*, y=1xy = \frac{1}{x}.
This guy has a point of symmetry at (0,0)(0, 0).
It's like a cool spinning top, balanced right at the origin!

STEP 4

Now, look at our *actual function*: f(x)=1x+3+2f(x) = \frac{1}{x+3} + 2.
See that +3+3 with the xx?
That means we **shift** the graph **3 units to the left**.
So, our *x*-value for the point of symmetry goes from 00 to 03=30 - 3 = \mathbf{-3}.

STEP 5

And what about that +2+2 hanging out at the end?
That **shifts** our graph **2 units up**!
So, our *y*-value for the point of symmetry goes from 00 to 0+2=20 + 2 = \mathbf{2}.

STEP 6

Putting it all together, our point of symmetry is at (3,2)(-3, 2)!
We shifted left by **3** and up by **2**, landing right on the spot!

STEP 7

The point of symmetry is (3,2)(-3, 2), so the answer is (d)!

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