Name:_________________________________________________Date:________________________Hour:___________

Unit 3: Quadratic Functions Review Key 1. The following functions represent three eggs that have been dropped off a roof, by a science class. In each case, describe what must have happened to make those equations model the fall of the egg. The first term controls the rate of falling. The second term controls whether or not there was an initial velocity (they threw it in the air before they dropped it) and the last term controls the height before the fall.

a.

𝑓(𝑥) = −16𝑥 2 + 2𝑥 + 15

The −16𝑥 2 is the part that is the gravity in the free fall. The 2𝑥 means that they must have thrown it up a rate of 2 (feet per second, probably) before they let it free fall. The 15 means that they were 15 (feet, probably) off the ground when they threw the egg. So they must have been standing 15ft off the ground and thrown the egg up at 2ft/sec before letting it free fall.

b. 𝑔(𝑥) = −16𝑥 2 + 12 The −16𝑥 2 is the part that is the gravity in the free fall. No middle term means that they must not have thrown it up before they let it free fall. The 12 means that they were 12 (feet, probably) off the ground when they threw the egg. So they must have been standing 12ft off the ground and dropped the egg, letting it free fall.

c. ℎ(𝑥) = −16𝑥 2 + 4𝑥 + 9 The −16𝑥 2 is the part that is the gravity in the free fall. The 4𝑥 means that they must have thrown it up a rate of 4 (feet per second, probably) before they let it free fall. The 9 means that they were 9 (feet, probably) off the ground when they threw the egg. So they must have been standing 9 ft off the ground and thrown the egg up at 4ft/sec before letting it free fall.

2. While standing at the top of a 49 foot ladder you drop a paintbrush and it freefalls to the ground. How many seconds will it take the paintbrush to hit the ground? Free falling objects can be modeled with 𝑦 = −16𝑥 2 + 𝑏𝑥 + 𝑐. The -16x2 is the pull of gravity in the free fall. The b is the initial velocity (whether or not it was thrown into the air before it fell) and the c is how high it was off the ground to start. There was no initial velocity (b) and it was 49 feet off the ground (c). So we model this situation with the equation 𝑦 = −16𝑥 2 + 49. If we want to know when the paintbrush hits the ground, then we want to know how many seconds (x) it takes for the height (y) to be 0. So we plug in 0 for y and then solve for x. 0 = −16𝑥 2 + 49 Try to get the x2 alone. −49 − 49 −49 = −16𝑥 2 ̅̅̅̅̅̅ ̅̅̅̅̅̅ −16 −16 3.0625 = 𝑥 2 To get the x squared undone we have to square root both sides. √3.0625 = √𝑥 2 √3.0625 = 𝑥 𝑎𝑛𝑑 − √3.0625 = 𝑥 Don’t forget that square rooting gives two answers! 1.75 = 𝑥 𝑎𝑛𝑑 − 1.75 = 𝑥 The x is counting seconds, so the negative answer doesn’t make any sense. So it takes 1.75 seconds for the paintbrush to hit the ground.

Name:_________________________________________________Date:________________________Hour:___________ 3. Is there an error in the solution? Explain your reasoning.

Yes, there is a mistake. Any time you take the square root to solve (like they did in the last step) you should get a positive answer and a negative answer. So their answer should be x=5 and x=5.

4. A storage container has the shape of a rectangular prism. Its height is 6 feet. Its length is two times its width. The volume is 288 cubic feet. Find the length and width of the container. The formula for the volume of a rectangular prism is Volume= lenghtxwidthxheight. We know the volume (that was given as 288 ft3) and we know the height (that was given as 6ft). We do not know the width, but we know that the length is 2 times whatever the width is. So we could call the width x, and call the length 2x. If we put in all of the pieces that we know then we will have the equation 288 = 2𝑥 ∙ 𝑥 ∙ 6. We can simplify that by combining the x’s and multiplying the numbers. 288 = 12𝑥 2 Now we just need to solve for x. First get the x2 alone. ̅̅̅̅ ̅̅̅̅ 12 12 24 = 𝑥 2 We square root to undo the square. Don’t forget that square rooting gives two answers. √24 = √𝑥 2 ±√24 = 𝑥 𝑜𝑟 ± 4.9 ≈ 𝑥 The negative answers won’t make any sense, so the width is √24 or about 4.9 feet.

5. Two students Two students are arguing about the quadratic function 𝑦 = 𝑥 2 + 6𝑥 − 3. Student 1 says the maximum value of the function is -3. Student 2 says that they know without even graphing it that there is no maximum value. Which student is correct? Explain your thinking. The number in front of the x2 is positive (it’s an invisible 1), which means that the parabola opens up. If it opens up then there will be no maximum value because it goes up and up forever and ever. It has a minimum value of -3, but no maximum value.

6. Consider the function 𝑓(𝑥) = −3𝑥 2 + 9𝑥 + 24. Which of the questions below has a different answer than the others? Circle the one that is different and then find “both” answers.

Name:_________________________________________________Date:________________________Hour:___________ “What is the maximum value of the function?” is asking for the highest y-value. “What is the greatest number in the range of the function?” is asking for the highest y-value. “What is the y-coordinate of the vertex of the graph is of the function?” is asking for the y-value of the highest point, or the highest y-value. “What is the axis of symmetry of the graph of the function?” is looking for the axis of symmetry for the function. That is the different question. The highest y-value: 30.8 The axis of symmetry: x=1.5 *To find these, graph the function and then use your calculator to find the maximum and the axis of symmetry.

7. If 𝑓(𝑥) = −16𝑡 2 + 48𝑡 + 32 models the fall of one object and 𝑔(𝑥) = −16𝑡 2 + 64𝑡 models the fall of another object, which object will reach the highest height? Explain. If I graph the two functions and then move the window around to fit the graph it looks like this: You can see in the graph that f(x) gets higher. Because it’s m maximum point is higher, it reaches the highest height.

8. Does one of the functions below have a higher y-intercept than the other? Explain.

X -3 -2 -1 0 1 2 3

Y 10 8 6 5 6 8 10

The y-intercept on the graph is where the graph crosses that x-axis. That is at 4 on the graph, so the graph has a y-intercept of 4. The y-intercept on the table is where the table has an xvalue of 0. That is where the y-value is 5, so the table has a y-intercept of 5. 5 is higher than 4, so the table has a higher y-intercept than the graph does.

9. Write a quadratic equation that has the same axis of symmetry as the function ℎ(𝑥) = 𝑥 2 + 4𝑥 + 11. Explain how you know that it does. If you graph the function and find the axis of symmetry you can see that the axis of symmetry is at -2. In order for a function to have an axis of symmetry at -2, the function needs to be shifted to the left 2. This is easiest to do in vertex form of the quadratic.

Name:_________________________________________________Date:________________________Hour:___________ The h controls the axis of symmetry, so as long as the h is -2, everything else can be whatever you want. Here are some things that would work: 𝑓(𝑥) = (𝑥 − −2)2 𝑔(𝑥) = 3(𝑥 + 2)2 − 7 𝑚(𝑥) = −5(𝑥 − −2)2 + 1 𝑘(𝑥) = (𝑥 + 2)2 − 4,329 These will all have an axis of symmetry at -2 because the h in the vertex form shifts the function to the left 2.

10. Write equation in vertex form that models the graph below. Vertex form of a quadratic function is 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 (see #9 for an explanation on vertex form). We can see the vertex on the graph at (-1,3), so we can put those in for h and k. 𝑦 = 𝑎(𝑥 − −1)2 + 3 or 𝑦 = 𝑎(𝑥 + 1)2 + 3 To find the a we need to plug in another point we know. The point (0,5) is on the graph, so we can plug in 0 for x and 5 for y to solve for a. 5 = 𝑎(0 + 1)2 + 3 Simplify what you can. 5 = 𝑎(1)2 + 3 5= 𝑎∙1+3 5 = 𝑎 + 3 Now get the a alone. −3 − 3 2 = 𝑎 so our a is 2. The whole equation must be 𝑦 = 2(𝑥 + 1)2 + 3

11. Write the equation in standard form that models the graph below. We need to write our answer in standard form, but it is easiest to write equations in vertex form, so we will start there. Vertex form of a quadratic function is 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 (see #9 for an explanation on vertex form). We can see the vertex on the graph at (-2,-7), so we can put those in for h and k. 𝑦 = 𝑎(𝑥 − −2)2 + −7 or 𝑦 = 𝑎(𝑥 + 2)2 − 7. To find the a we need to plug in another point we know. The point (0,-3) is on the graph, so we can plug in 0 for x and -3 for y to solve for a. −3 = 𝑎(0 + 2)2 − 7 Simplify. −3 = 𝑎(2)2 − 7 −3 = 𝑎 ∙ 4 − 7 −3 = 4𝑎 − 7 Get the a alone. +7 +7 4 = 4𝑎 4̅ 4̅ 1 = 𝑎 so our a is 1. The whole equation must be 𝑦 = 1(𝑥 + 2)2 − 7 or 𝑦 = (𝑥 + 2)2 − 7. We were supposed to write it in standard form, so we need to simplify the vertex form down to standard form. 𝑦 = (𝑥 + 2)2 − 7 The 2 means that we have (x+2) twice being multiplied. 𝑦 = (𝑥 + 2)(𝑥 + 2) − 7 We multiply everything in the first parenthesis by everything in the second parenthesis. 𝑦 = 𝑥 2 + 2𝑥 + 2𝑥 + 4 − 7 Combine all the like terms. 𝑦 = 𝑥 2 + 4𝑥 − 3 Done!

Name:_________________________________________________Date:________________________Hour:___________ 12. Write the equation in vertex form that models the table below.

X 0 1 2 3 4

Y -10 -13 -14 -13 -10

Vertex form of a quadratic function is 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 (see #9 for an explanation on vertex form). The vertex is the turning point, so we can see on the table that (2,-14) must be the vertex, because that is where the graph turns around and starts repeating values. That will be our h and k. 𝑦 = 𝑎(𝑥 − 2)2 + −14 or 𝑦 = 𝑎(𝑥 − 2)2 − 14. To find the a we need to plug in an x and a y and then solve for a. We can use any of the points on the table as our x and y. I am going to use the point (0,-10).

−10 = 𝑎(0 − 2)2 − 14 Simplify. −10 = 𝑎(−2)2 − 14 −10 = 𝑎 ∙ 4 − 14 −10 = 4𝑎 − 14 Get the a alone. +14 + 14 4 = 4𝑎 4̅ 4̅ 1 = 𝑎 So our a is 1. So the equation must be 𝑦 = 1(𝑥 − 2)2 − 14 or 𝑦 = (𝑥 − 2)2 − 14.

13. Write the equation in standard form that models the table below.

X -4 -3 -2 -1 0

Y -3 -6 -3 6 21

This equation needs to be written in standard form, but vertex form makes it easier to see the pieces, so we will write in vertex form first and then simplify it. Vertex form of a quadratic function is 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 (see #9 for an explanation on vertex form). The vertex is the turning point, so we can see on the table that (-3,-6) must be the vertex, because that is where the graph turns around and starts repeating values. That will be our h and k. 𝑦 = 𝑎(𝑥 − −3)2 + −6 or 𝑦 = 𝑎(𝑥 + 3)2 − 6.

To find the a we need to plug in an x and a y and then solve for a. We can use any of the points on the table as our x and y. I am going to use the point (0,21). 21 = 𝑎(0 + 3)2 − 6 21 = 𝑎(3)2 − 6 21 = 𝑎 ∙ 9 − 6 21 = 9𝑎 − 6 Now get the a alone. +6 +6 27 = 9𝑎 9̅ 9̅ 3 = 𝑎 so our a is 3. So the equation must be 𝑦 = 3(𝑥 + 3)2 − 6. We are supposed to write it in standard form though, so we need to simplify it. 𝑦 = 3(𝑥 + 3)2 − 6 The 2 means we have (x+3) twice. 𝑦 = 3(𝑥 + 3)(𝑥 + 3) − 6 Let’s do the multiplication first, times everything in the first polynomial by everything in the second polynomial. 𝑦 = 3(𝑥 2 + 3𝑥 + 3𝑥 + 9) − 6 Now multiply everything by 3. 𝑦 = 3𝑥 2 + 9𝑥 + 9𝑥 + 27 − 6 Now combine all of the like terms.

Name:_________________________________________________Date:________________________Hour:___________ 𝑦 = 3𝑥 2 + 18𝑥 + 21 Done!

14. Is the function 𝑦 = 𝑥 3 + 3𝑥 2 + 6 quadratic? Explain how you know. No, we know it is not quadratic because quadratic functions are functions with an x 2. Anything else is not quadratic.

15. Student 1 models a quadratic situation with the equation 𝑓(𝑥) = 𝑥 2 − 10𝑥 + 18. Student 2 models the same situation with the equation 𝑦 = (𝑥 − 5)2 + 23. Is it possible that both students are correct? Explain your thinking. It is possible that both students are correct because student 1 has their answer in standard form and student 2 has their answer in vertex form. If we simplify Student 2’s work and put it in standard form, we will be able to see if it is like student 1’s. 𝑦 = (𝑥 − 5)2 + 23 The 2 means we have (x-5) twice. 𝑦 = (𝑥 − 5)(𝑥 − 5) + 23 Do the multiplication by multiplying everything in the first polynomial by everything in the second polynomial. 𝑦 = 𝑥 2 − 5𝑥 − 5𝑥 + 25 + 23 Now combine the like terms. 𝑦 = 𝑥 2 − 10𝑥 + 47 . This is still not the same as student 2’s, so it is not possible that both students are correct because they have different answers.