Algebra 2 10 Questions Solve.

Question

1. A parabola has an axis of symmetry y = -4 and passes through the point (2, -1). Find another point that lies on the graph of the parabola.

2. Let the graph of g be a horizontal shrink by a factor of 1/3, followed by a translation 1 unit up of the graph of f(x) = x2. Write a rule for g.

3. Let the graph of g be a translation 2 units up and 2 units right, followed by a reflection in the y-axis of the graph of f(x) = -(x + 3)2. Write a rule for g.

4. Identify the focus, directrix, and axis of symmetry of x = -(1/20)y2.

Write a rule for g and identify the vertex.

5. Let g be a translation 2 units up, followed by a reflection in the x-axis and a vertical stretch by a factor of 4 of the graph of f(x) = x2

6. Let g be a horizontal shrink by a factor of 1/3, followed by a translation 2 units up and 4 units left of the graph of f(x) = (3x - 2)2 + 5

Find the x-intercepts of the graph of the function. Then describe where the function is increasing and decreasing.

7. g(x) = -1(x - 4)(x + 2)

8. g(x) = (1/4)(x - 6)(x - 3)

9. An object is launched directly overhead at 36 meters per second. The height (in meters) of the object is given by h(t) = -16t2 + 36t + 5, where t is the time (in seconds) since the object was launched. For how many seconds is the object at or above a height of 25 meters?

10. A model rocket is launched from the top of a building. The height (in meters) of the rocket above the ground is given by h(t) = -6t2 +24t +14, where t is the time (in seconds) since the rocket was launched. What is the rocket’s maximum height?

Answer 1

y=-4 is the axis of symmetry

a point (2, -1) is on the horizontal parabola

another point on the parabola is (2, -7)

because -1- -4 = 3 and -4- -7 = 3

those two points are symmetric about the line y=-4

(2,-1) is 3 units above y=-4, and (2,-7) is 3 units below y=-4

jumping to the last question, number 10

h(t) =-6t^2 +24t +14

take the derivative and set = 0

h'(t) = -12t +24=0

t = 24/12 = 2 seconds to reach max height

max height = h(2) =-6(2)^2 +24(2)+14 = -24+48+14= 38 meters high

but you're listed under algebra 2, not calculus,

so you may or may not yet understand the above

solution, but you now at least know the answer, h=38

in algebra, you need to rewrite h(t) in vertex form,

then the y coordinate of the vertex = the max height

-6t^2+24t+14

complete the square, add and subtract the same value of 24

= -6(t^2-4t +4) +14 +(4)(6)

=-6(t-2)^2+38

vertex form is a(t-h)^2 +k

where (h,k)= vertex= (2, 38) h=2 seconds, k=38 meters = max height