Math / Algebra

QuestionFor the quadratic function $f(x)=x^{2}+6 x$ parts (a) through (f). (Type your answer in interval notation.) The range of $f$ is $(-9, \infty)$ . (Type your answer in interval notation.) (e) Determine where the quadratic functio increasing and where it is decreasing.
The function is increasing on the interval (Type your answer in interval notation.)

Studdy Solution

STEP 1

What is this asking? We're given a parabola $f(x) = x^2 + 6x$ and we need to find where it's going uphill and downhill! Watch out! Remember increasing and decreasing intervals are about the * $\ x$ * values, not the * $\ y$ * values!

STEP 2

1. Find the vertex
2. Determine increasing and decreasing intervals

STEP 3

Let's complete the square to put our parabola in vertex form.
This makes it super easy to see the vertex, which is like the tip of the parabola.
We have $f(x) = x^2 + 6x$ .
To complete the square, we take half of the coefficient of the * $\ x$ * term, which is $\frac{6}{2} = \textbf{3}$ , and square it: $\textbf{3}^2 = \textbf{9}$ .
So, we add and subtract 9:
$f(x) = x^2 + 6x + 9 - 9$

STEP 4

Now, we can rewrite the first three terms as a perfect square:
$f(x) = (x+3)^2 - 9$ This is vertex form, and it tells us the vertex is at $(-3, -9)$ .
The * $\ x$ *-coordinate of the vertex is $\textbf{-3}$ .

STEP 5

Since the coefficient of $x^2$ is positive ( $1$ ), our parabola opens upwards, like a smile!
This means it's decreasing to the left of the vertex and increasing to the right of the vertex.

STEP 6

The function is decreasing on the interval $(-\infty, \textbf{-3})$ and increasing on the interval $(\textbf{-3}, \infty)$ .
Remember, we're talking about * $\ x$ * values here!

STEP 7

The function is increasing on the interval $(\textbf{-3}, \infty)$ .

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Studdy Solution

STEP 1

What is this asking? We're given a parabola $f(x) = x^2 + 6x$ and we need to find where it's going uphill and downhill! Watch out! Remember increasing and decreasing intervals are about the * $\ x$ * values, not the * $\ y$ * values!

STEP 2

1. Find the vertex
2. Determine increasing and decreasing intervals

STEP 3

STEP 4

Now, we can rewrite the first three terms as a perfect square:
$f(x) = (x+3)^2 - 9$ This is vertex form, and it tells us the vertex is at $(-3, -9)$ .
The * $\ x$ *-coordinate of the vertex is $\textbf{-3}$ .

STEP 5

Since the coefficient of $x^2$ is positive ( $1$ ), our parabola opens upwards, like a smile!
This means it's decreasing to the left of the vertex and increasing to the right of the vertex.

STEP 6

The function is decreasing on the interval $(-\infty, \textbf{-3})$ and increasing on the interval $(\textbf{-3}, \infty)$ .
Remember, we're talking about * $\ x$ * values here!

STEP 7

The function is increasing on the interval $(\textbf{-3}, \infty)$ .

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Studdy Solution

STEP 1

What is this asking? We're given a parabola f(x)=x2+6xf(x) = x^2 + 6xf(x)=x2+6x and we need to find where it's going uphill and downhill! Watch out! Remember increasing and decreasing intervals are about the * x\ x x* values, *not* the * y\ y y* values!

STEP 2

1. Find the vertex2. Determine increasing and decreasing intervals

STEP 3

STEP 4

Now, we can rewrite the first three terms as a perfect square:f(x)=(x+3)2−9f(x) = (x+3)^2 - 9f(x)=(x+3)2−9This is **vertex form**, and it tells us the **vertex** is at (−3,−9)(-3, -9)(−3,−9).The * x\ x x*-coordinate of the vertex is -3\textbf{-3}-3.

STEP 5

Since the coefficient of x2x^2x2 is **positive** (111), our parabola opens **upwards**, like a smile!This means it's decreasing to the *left* of the vertex and increasing to the *right* of the vertex.

STEP 6

The function is **decreasing** on the interval (−∞,-3)(-\infty, \textbf{-3})(−∞,-3) and **increasing** on the interval (-3,∞)(\textbf{-3}, \infty)(-3,∞).Remember, we're talking about * x\ x x* values here!

STEP 7

The function is increasing on the interval (-3,∞)(\textbf{-3}, \infty)(-3,∞).

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What is this asking? We're given a parabola $f(x) = x^2 + 6x$ and we need to find where it's going uphill and downhill! Watch out! Remember increasing and decreasing intervals are about the * $\ x$ * values, not the * $\ y$ * values!

1. Find the vertex
2. Determine increasing and decreasing intervals

Now, we can rewrite the first three terms as a perfect square:
$f(x) = (x+3)^2 - 9$ This is vertex form, and it tells us the vertex is at $(-3, -9)$ .
The * $\ x$ *-coordinate of the vertex is $\textbf{-3}$ .

Since the coefficient of $x^2$ is positive ( $1$ ), our parabola opens upwards, like a smile!
This means it's decreasing to the left of the vertex and increasing to the right of the vertex.

The function is decreasing on the interval $(-\infty, \textbf{-3})$ and increasing on the interval $(\textbf{-3}, \infty)$ .
Remember, we're talking about * $\ x$ * values here!

The function is increasing on the interval $(\textbf{-3}, \infty)$ .