Quadratic Summary
Standard Form of a Quadratic:
y=ax^2 +bx+c
Coefficient: a,b
Constant: c
Example:
y=x^2+5x+3
Coefficient: a=1, b=5
Constant: c=3
To find the vertex of an equation using standard form, you can plug in x values into the equation, solve, and graph your x and y value solutions. You can continue to do this until you start forming a parabola. An easy way to compile your data is by using an x and y table. Although finding the vertex using standard form is a longer process, it can still get you the correct answer.
So if you take the equation y=x^2+5x+3 and make a x and y table in order to eventually find the vertex, the first few point would look something like this:
y=ax^2 +bx+c
Coefficient: a,b
Constant: c
- Standard form is used to figure out what your x and y intercepts are.
- It is also used to turn the equation into different forms
- What makes this the equation for the standard form of a quadratic is the power of two on the x value
Example:
y=x^2+5x+3
Coefficient: a=1, b=5
Constant: c=3
To find the vertex of an equation using standard form, you can plug in x values into the equation, solve, and graph your x and y value solutions. You can continue to do this until you start forming a parabola. An easy way to compile your data is by using an x and y table. Although finding the vertex using standard form is a longer process, it can still get you the correct answer.
So if you take the equation y=x^2+5x+3 and make a x and y table in order to eventually find the vertex, the first few point would look something like this:
You would continue your chart in order to find more x and y values, and graph them as you go. Eventually you will find your vertex when your points start to look like a parabola.
If you graph this equation, this is what you get:
When you change the a coefficient (1) to a negative, the graph opens to the bottom:
If you change the a coefficient to a different number, in this case you could change 1 to 3, your vertex changes. The parabola also gets thinner:
If you change the a coefficient to a negative number, going from 1 to -3, your vertex also changes, and the parabola gets thinner, but it opens to the bottom:
If you change the b coefficiant to a negiative, going from 5 to -5, your parabola shifts in the positive x direction:
If you change the c constant to a negative, the parabola will move down the y axis:
Vertex form of a Quadratic:
y= a(x-h)^2+k
Vertex: h,k
Example:
y=(x-2)^2+5
To graph a quadratic using vertex form, first you identify the vertex. To find the vertex is very simple, it will always be (h, k). But you take the opposite of h, so in this case your vertex would be (-2, 5).
y=(x-2)^2+5
h, k= 2, 5
Then find the y intercept. To find it, make x equal to zero and solve.
y=(x-2)^2+5
y=(0-2)^2+5
y=(-2)^2+5
y=4+5
y=9
To find the x intercepts, let y be zero and solve using the square root principle.
Then, you should be able to graph your quadratic. Your graph would look something like this:
Vertex: h,k
- Vertex form is an easy way to find out what the vertex of an equation is just by looking at it
- Its a faster way to graph quadratics
Example:
y=(x-2)^2+5
To graph a quadratic using vertex form, first you identify the vertex. To find the vertex is very simple, it will always be (h, k). But you take the opposite of h, so in this case your vertex would be (-2, 5).
y=(x-2)^2+5
h, k= 2, 5
Then find the y intercept. To find it, make x equal to zero and solve.
y=(x-2)^2+5
y=(0-2)^2+5
y=(-2)^2+5
y=4+5
y=9
To find the x intercepts, let y be zero and solve using the square root principle.
Then, you should be able to graph your quadratic. Your graph would look something like this:
X Intercepts (Roots):
The x intercepts of a quadratic are the places that the parabola crosses the x axis
And easy way to find the x intercepts of a quadratic is by factoring.
Example:
y= x^2+ 11k+28
First, set the equation equal to 0
0= x^2+ 11k+28
Then, solve for x by factoring
x^2+ 11k+28
(x+4)(x+7)=0
x+4=0
x=-4
x+7=0
x=-7
Solution: (-4,-7)
If you want to check your equation, plug the x values back into the equation and solve.
And easy way to find the x intercepts of a quadratic is by factoring.
Example:
y= x^2+ 11k+28
First, set the equation equal to 0
0= x^2+ 11k+28
Then, solve for x by factoring
x^2+ 11k+28
(x+4)(x+7)=0
x+4=0
x=-4
x+7=0
x=-7
Solution: (-4,-7)
If you want to check your equation, plug the x values back into the equation and solve.
Converting from Standard form to Vertex form and Vertex form to Standard form (and challenge option #2):
You can convert standard form and vertex form back and forth and get the original/correct answer
Vertex form to standard form:
Vertex form= y= (x-h)^2+k
Standard form= y=ax^2 +bx+c
y= (x-h)^2+k
Factor the equation to get it into standard form
(x-h)(x-h)+k
Solution: x^2-h+k
Standard form to Vertex form:
Vertex form= y= (x-h)^2+k
Standard form= y=ax^2 +bx+c
y=ax^2 +bx+c
Use the FOIL method
(ax+b)(ax+b)
b x b = b^2
b^2 + b = k
y= (x-h)^2+k
b= h
c-h^2= k
Examples:
Vertex to Standard:
Solution: x^2-6x-3
1+2+3 . (x-h)(x-h)+k
4. x^2-6x-3
Standard to Vertex:
Solution: (x-3)^2-12
2. (ax+b)(ax+b)
3. b x b = b^2
4. b^2 + b = k
5. y= (x-h)^2+k
b= h
c-h^2= k
Vertex form to standard form:
Vertex form= y= (x-h)^2+k
Standard form= y=ax^2 +bx+c
y= (x-h)^2+k
Factor the equation to get it into standard form
(x-h)(x-h)+k
Solution: x^2-h+k
Standard form to Vertex form:
Vertex form= y= (x-h)^2+k
Standard form= y=ax^2 +bx+c
y=ax^2 +bx+c
Use the FOIL method
(ax+b)(ax+b)
b x b = b^2
b^2 + b = k
y= (x-h)^2+k
b= h
c-h^2= k
Examples:
Vertex to Standard:
- (x-3)^2-12
- (x-3)(x-3)
- x^2-6x+9-12
- x^2-6x-3
Solution: x^2-6x-3
1+2+3 . (x-h)(x-h)+k
4. x^2-6x-3
Standard to Vertex:
- x^2-6x-3
- (x-3)(x-3)
- -3 x -3 = -9
- -9 - -3 = -12
- (x-3)^2-12
Solution: (x-3)^2-12
2. (ax+b)(ax+b)
3. b x b = b^2
4. b^2 + b = k
5. y= (x-h)^2+k
b= h
c-h^2= k
Blocks Again (Challenge Option):
1. What is the growing pattern here?
The pattern grows one block from eight different directions each stage, two going down, three diagonally down to the right, and three to the right. As shown below, each stage it adds one more layer. The constant shape/tiles is represented by the orange. The constant stays the same the whole time. The blue and purple represent the growth.
The pattern grows one block from eight different directions each stage, two going down, three diagonally down to the right, and three to the right. As shown below, each stage it adds one more layer. The constant shape/tiles is represented by the orange. The constant stays the same the whole time. The blue and purple represent the growth.
2. Sketch what the next few figures should look like
Below I sketched what figures 1, 2, 3, 4, 5, 6, 7, and 8 look like.
Below I sketched what figures 1, 2, 3, 4, 5, 6, 7, and 8 look like.
3. Using what you have discovered thus far can you figure out how many blocks would be in Figure 50? Figure 100?
You could always draw all the stages up to those points and count all the blocks, but that would take way too long. So what I did was look back on the first Growing Tiles problem. I read over my work and relearned the problem, and I tried to use the equation that I had used for the growing tiles problem. When I inputted all the correct variables though, for some reason I didn't get the correct answer. I tried creating an entirely new equation as well, but after many failed attempts I still could't come up with an answer. The day after attempting this problem for the first day I went and asked some of my classmates if they figured out how to do the problem, and most of them hadn't even tried to do it. As for the people who did, they couldn't come up with a conclusion either. I am hoping to spend some more time with this problem later and see if I can figure it out at my own pace.
4) Using what you have discovered thus far can you come up with an equation that determines the number of blocks for a given Figure?
Unfortunately I couldn't come up with an equation that fit this problem, I did try using the equation that worked flawlessly with the other growing tiles problem. There must be another variable that I am not accounting for, and I hope to eventually complete this problem.
You could always draw all the stages up to those points and count all the blocks, but that would take way too long. So what I did was look back on the first Growing Tiles problem. I read over my work and relearned the problem, and I tried to use the equation that I had used for the growing tiles problem. When I inputted all the correct variables though, for some reason I didn't get the correct answer. I tried creating an entirely new equation as well, but after many failed attempts I still could't come up with an answer. The day after attempting this problem for the first day I went and asked some of my classmates if they figured out how to do the problem, and most of them hadn't even tried to do it. As for the people who did, they couldn't come up with a conclusion either. I am hoping to spend some more time with this problem later and see if I can figure it out at my own pace.
4) Using what you have discovered thus far can you come up with an equation that determines the number of blocks for a given Figure?
Unfortunately I couldn't come up with an equation that fit this problem, I did try using the equation that worked flawlessly with the other growing tiles problem. There must be another variable that I am not accounting for, and I hope to eventually complete this problem.
Personal Growth Reflection:
I honestly think I have done outstanding during this unit. I have made sure to focus every single day in class and I fully understand every concept we went over these past weeks, and I am very confident in my ability to even teach others about what we've learned. I give my all in class every day, and I'm really hoping to continue this attitude throughout the rest of the school year, at least to the best of my ability. I have used several habits of a mathematician this unit, including being confident, patient, and persistent. I think in the beginning of this unit I was really unsure about my math skills, but going up to the board and sharing my work with others has boosted my confidence and now I can share ideas with the entire class even if I know they aren't right. I also try to explore problems and take any challenge option that is given to me, even if I'm unsure how to even attempt it. Another habit of a mathematician I used was looking for patterns. This unit especially has made me utilize that skill, like in the growing tiles problem and even with quadratics. This has also made me talk to my classmates because sometimes its harder to recognize patterns, but when you ask other people what they think, they open up your mind to look at the problem in a different way. The problem is that I am an independent student and I usually don't need help from others with things like quadratics. So with the next unit I'm hoping to change that and collaborate with others more and ask them about their ideas. All in all, I think this was a very successful couple weeks especially for me. But the challenge will be to continue it, and that will be my goal for the upcoming months.