Math  /  Algebra

QuestionFind the vertex and axis of the graph of the function. f(x)=2(x3)2+9f(x)=2(x-3)^{2}+9
The vertex is \square (Type an ordered pair.)

Studdy Solution

STEP 1

What is this asking? We're looking for the **lowest point** on the curve of this *parabola* and the **vertical line** that goes right through it! Watch out! Don't mix up the xx and yy values of the vertex!

STEP 2

1. Recall the vertex form.
2. Identify the vertex.
3. Find the axis.

STEP 3

The **vertex form** of a parabola's equation is f(x)=a(xh)2+kf(x) = a(x - h)^2 + k, where (h,k)(h, k) is the **vertex** of the parabola.
Remember, the **vertex** is like the tip of the parabola, either the very top or the very bottom!

STEP 4

Let's **compare** our given function, f(x)=2(x3)2+9f(x) = 2(x - 3)^2 + 9, to the **vertex form**, f(x)=a(xh)2+kf(x) = a(x - h)^2 + k.

STEP 5

We can see that a=2a = 2, h=3h = 3, and k=9k = 9.
This tells us that the **vertex** of our parabola is at the point (3,9)(3, 9)!

STEP 6

The **axis of symmetry** is a *vertical line* that goes right through the **vertex**.
Since it's vertical, its equation will always look like x=some numberx = \text{some number}.

STEP 7

Since our **vertex** is at (3,9)(3, 9), the xx-value of our vertex is 33.
This means the **axis of symmetry** is the line x=3x = 3.

STEP 8

The **vertex** is (3,9)(3, 9) and the **axis of symmetry** is x=3x = 3.

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