What is the range of y = –x2+2x+8?

All real numbers greater than or less than 9.

Step-by-step explanation:  The range is possible y-values. In a parabola the range starts at a particular place then it either goes up or goes down. Since the leading factor has a negative in front of it, the parabola will open downward meaning most likely all y-values will be lest than the include max point. Now let find the max point using the formula -b/2a then substituting that into the parabola. -2/2(-1)= 1, then plug it in. y=-(1^2)+2(1)+8 which equal -1+2+8 which equal 9 so the max point is 9. Since the graph open downward all the points that are possible will either be at 9 or below it so the range is All  real numbers greater than or less than 9. or (-∞,∞)∪(∞,9]

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true