Math / Algebra

QuestionPart 11 of 11
For the quadratic function $f(x)=x^{2}+2 x+1$ , answer parts (a) through ( $f$ ). (Use integers or fractions for any numbers in the equation.) Is the graph concave up or concave down? Concave down Concave up (b) Find the $y$ -intercept and the $x$ -intercepts, if any. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The $x$ -intercept(s) is/are -1 . $\square$ : (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) B. There are no $x$ -intercepts.
What is the y-intercept? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The y-intercept is 1. $\square$ (Type an integer or a simplified fraction.) B. There is no $y$ -intercept. (c) Use parts (a) and (b) to graph the function.
Use the graphing tool to graph the function. $\square$ (d) Find the domain and the range of the quadratic function.
The domain of $f$ is $(-\infty, \infty)$ . (Type your answer in interval notation.) The range of $f$ is $[0, \infty)$ . (Type your answer in interval notation.) (e) Determine where the quadratic function is increasing and where it is decreasing.
The function is increasing on the interval $(-1, \infty)$ . (Type your answer in interval notation.) The function is decreasing on the interval $(-\infty,-1)$ . (Type your answer in interval notation.) (f) Determine where $f(x)>0$ and where $f(x)<0$ . Select the correct choice below and fill in the answer box(es) within your choice. (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.) A. $f(x)>0$ on and $f(x)<0$ on $\square$ B. $f(x)<0$ on and $f(x)$ is never positive $\square$ C. $f(x)>0$ on $(-\infty,-1) \cup(-1, \infty)$ and $f(x)$ is never negative ctbook Ask my instructor

Studdy Solution

STEP 1

1. We are analyzing the quadratic function $f(x) = x^2 + 2x + 1$ .
2. We need to determine the concavity, intercepts, graph, domain, range, intervals of increase/decrease, and where the function is positive or negative.

STEP 2

1. Determine the concavity of the graph.
2. Find the $y$ -intercept and $x$ -intercepts.
3. Use the information to graph the function.
4. Determine the domain and range of the function.
5. Identify intervals where the function is increasing or decreasing.
6. Determine where $f(x) > 0$ and $f(x) < 0$ .

STEP 3

The quadratic function $f(x) = x^2 + 2x + 1$ has a positive leading coefficient (1), indicating that the graph is concave up.

STEP 4

To find the $y$ -intercept, set $x = 0$ : $f(0) = 0^2 + 2 \times 0 + 1 = 1$ Thus, the $y$ -intercept is $(0, 1)$ .
To find the $x$ -intercepts, set $f(x) = 0$ : $x^2 + 2x + 1 = 0$ Factor the quadratic: $(x+1)^2 = 0$ So, $x = -1$ . Thus, the $x$ -intercept is $(-1, 0)$ .

STEP 5

Use the intercepts and concavity to sketch the graph. The vertex is at $(-1, 0)$ and the parabola opens upwards.

STEP 6

The domain of any quadratic function is all real numbers: $(-\infty, \infty)$
The range, since the parabola opens upwards and the vertex is at $y = 0$ , is: $[0, \infty)$

STEP 7

The function is decreasing on the interval $(-\infty, -1)$ and increasing on the interval $(-1, \infty)$ .

STEP 8

Since the vertex is at $(-1, 0)$ and the parabola opens upwards, $f(x) > 0$ on $(-\infty, -1) \cup (-1, \infty)$ and $f(x)$ is never negative.
The function is concave up, the $y$ -intercept is $(0, 1)$ , the $x$ -intercept is $(-1, 0)$ , the domain is $(-\infty, \infty)$ , the range is $[0, \infty)$ , it is increasing on $(-1, \infty)$ , decreasing on $(-\infty, -1)$ , and $f(x) > 0$ on $(-\infty, -1) \cup (-1, \infty)$ .

Was this helpful?

Studdy Solution

STEP 1

1. We are analyzing the quadratic function $f(x) = x^2 + 2x + 1$ .
2. We need to determine the concavity, intercepts, graph, domain, range, intervals of increase/decrease, and where the function is positive or negative.

STEP 2

STEP 3

The quadratic function $f(x) = x^2 + 2x + 1$ has a positive leading coefficient (1), indicating that the graph is concave up.

STEP 4

To find the $y$ -intercept, set $x = 0$ : $f(0) = 0^2 + 2 \times 0 + 1 = 1$ Thus, the $y$ -intercept is $(0, 1)$ .
To find the $x$ -intercepts, set $f(x) = 0$ : $x^2 + 2x + 1 = 0$ Factor the quadratic: $(x+1)^2 = 0$ So, $x = -1$ . Thus, the $x$ -intercept is $(-1, 0)$ .

STEP 5

Use the intercepts and concavity to sketch the graph. The vertex is at $(-1, 0)$ and the parabola opens upwards.

STEP 6

The domain of any quadratic function is all real numbers: $(-\infty, \infty)$
The range, since the parabola opens upwards and the vertex is at $y = 0$ , is: $[0, \infty)$

STEP 7

The function is decreasing on the interval $(-\infty, -1)$ and increasing on the interval $(-1, \infty)$ .

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. We are analyzing the quadratic function f(x)=x2+2x+1 f(x) = x^2 + 2x + 1 f(x)=x2+2x+1.2. We need to determine the concavity, intercepts, graph, domain, range, intervals of increase/decrease, and where the function is positive or negative.

STEP 2

STEP 3

The quadratic function f(x)=x2+2x+1 f(x) = x^2 + 2x + 1 f(x)=x2+2x+1 has a positive leading coefficient (1), indicating that the graph is concave up.

STEP 4

STEP 5

Use the intercepts and concavity to sketch the graph. The vertex is at (−1,0) (-1, 0) (−1,0) and the parabola opens upwards.

STEP 6

The domain of any quadratic function is all real numbers: (−∞,∞) (-\infty, \infty) (−∞,∞)The range, since the parabola opens upwards and the vertex is at y=0 y = 0 y=0, is: [0,∞) [0, \infty) [0,∞)

STEP 7

The function is decreasing on the interval (−∞,−1) (-\infty, -1) (−∞,−1) and increasing on the interval (−1,∞) (-1, \infty) (−1,∞).

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We are analyzing the quadratic function $f(x) = x^2 + 2x + 1$ .
2. We need to determine the concavity, intercepts, graph, domain, range, intervals of increase/decrease, and where the function is positive or negative.

The quadratic function $f(x) = x^2 + 2x + 1$ has a positive leading coefficient (1), indicating that the graph is concave up.

Use the intercepts and concavity to sketch the graph. The vertex is at $(-1, 0)$ and the parabola opens upwards.

The domain of any quadratic function is all real numbers: $(-\infty, \infty)$
The range, since the parabola opens upwards and the vertex is at $y = 0$ , is: $[0, \infty)$

The function is decreasing on the interval $(-\infty, -1)$ and increasing on the interval $(-1, \infty)$ .