Math  /  Algebra

QuestionFor the quadratic function f(x)=x2+2x+1f(x)=x^{2}+2 x+1, answer parts (a) through ( ff ). (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) B. There are no xx-intercepts.
What is the yy-intercept? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The yy-intercept is 1 . (Type an integer or a simplified fraction.) B, There is no yy-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 ff is (,)(-\infty, \infty). (Type your answer in interval notation.) The range of ff is [0,)[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,)(-1, \infty). (Type your answer in interval notation.) The function is decreasing on the interval (,1)(-\infty,-1). (Type your answer in interval notation.) (f) Determine where f(x)>0f(x)>0 and where f(x)<0f(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)>0f(x)>0 on \square and f(x)<0f(x)<0 on \square \square B. f(x)<0f(x)<0 on \square and f(x)f(x) is never positive \square C. f(x)>0f(x)>0 on \square and f(x)f(x) is never negative

Studdy Solution
Since the vertex is at (1,0) (-1, 0) and the parabola opens upwards, f(x) f(x) is never negative.
f(x)>0 on (,1)(1,) f(x) > 0 \text{ on } (-\infty, -1) \cup (-1, \infty)
f(x)=0 at x=1 f(x) = 0 \text{ at } x = -1
Thus, the correct choice is:
C. f(x)>0 on (,1)(1,) f(x) > 0 \text{ on } (-\infty, -1) \cup (-1, \infty) and f(x) f(x) is never negative.

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