Math  /  Algebra

QuestionFor the quadratic function f(x)=x2+2x+1f(x)=x^{2}+2 x+1, answer parts (a) through ( ff ).
Is the graph concave up or concave down?
Concave down (Type your answer in interval notation.) (f) Determine where f(x)>0f(x)>0 and where f(x)<0f(x)<0. Select the correct cholce 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)>0f(x)>0 on \square and f(x)<0f(x)<0 on \square B. f(x)<0f(x)<0 on 0 and f(x)f(x) is never positive C. f(x)>0f(x)>0 on \square and f(x)f(x) is never negative

Studdy Solution
Since the vertex (1,0)(-1, 0) is the minimum point and the graph is concave up, f(x)0 f(x) \geq 0 for all x x .
The function f(x)=0 f(x) = 0 only at the vertex x=1 x = -1 .
Thus, f(x)>0 f(x) > 0 for all x1 x \neq -1 .
Therefore, the correct choice is: C. f(x)>0 f(x) > 0 on (,1)(1,)(-\infty, -1) \cup (-1, \infty) and f(x) f(x) is never negative.
The solution is: C. f(x)>0 f(x) > 0 on (,1)(1,)(-\infty, -1) \cup (-1, \infty) and f(x) f(x) is never negative.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

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

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord