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Google Slide

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I first started off by sketching the image of my guitar and then placing a grid right over it. From there I figured out the x and y intercepts. The x-intercepts were (-1, 0), (0, 0), and (1, 0). The y-intercept was (0, 0). From the intercepts I was able to write the equation: -(x-0)(x+0)(x+1)(x-1). I solved that to get the equation -x^4+x^2. The domain and range are both all real numbers, in other words, (-infinity, infinity).

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Shown here is the graph of my polynomial function, the equation, the x and y intercepts, and the domain and range.

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The notes of the song on the staff visually create a polynomial function.

The polynomial is 7th degree

The domain and range are both all real numbers

If the treble clef is considered 0, the first note is 1, and the second note is 2 (and so on), then the decreasing intervals would be (-∞,3) (3,6) (9,12) (14,∞) and the increasing intervals would be (6,9) (12,14)

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Polynomial
Curved lines, the domain and range
running through nature

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I traced the photo, on the line of my cat's spine at an angle, and traced that onto a graph, to get intercepts at X = -4 and at X = 0 . since the function of his spine projected onto a 2 dimensional surface is a 3rd degree polynomial function, all that had to be added to my  (x+4) was a x^2, and to make the equation negative, in order to reflect it over the horizontal x axis, making the equation f(x) = -x^3-4x^2. since this equation goes in both directions on both axes to infinity the domain and range are both, (-inf, inf) .

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Included are a graph, intercepts, an equation in two forms, and the dominant and range of that function

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Included are a graph, intercepts, an equation in two forms, and the dominant and range of that function

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Mason Diamond

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The first thing that I did was sketch a zoomed in part of my photo. I then found the x intercepts which were (-3.5,0) (-1.5,0) (1.5,0) and (3,0). The next thing I did was find the y intercept which was (0,2). I then put the intercepts into an equation which was f(x)=a(x+3.5)(x+1.5)(x-1.5)(x-3). I plugged in my y intercept of (0,2) and from this equation I found a=16/189. I then plugged that in for a and had an equation of f(x)=(16/189)(x+3.5)(x+1.5)(x-1.5)(x-3). I also found the domain and range of the function which were both (-infinity,infinity) because the x and y axis can go for as long as possible.

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The photo includes the graph, x and y intercepts, the domain and range, and the equation of the function.

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I first started by sketching the picture of my mountain on a piece of graph paper. The points on the graph that were easy for me to use were (-5,0), (0,7), (5,5), (10,2), and (12,0). I used the method you taught us in class to determine the equation of the polynomial. My X and Y intercepts were all visible from the graph I drew (X ints: (-5,0), (12,0). Y int: (7,0)).

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The graph, equation, domain/range, X & Y intercepts, and end behavior

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I started off by using a website to place a grid over my image. Then, I used the grid to find eight points along the line. I made sure that these points included all x-intercepts, y-intercepts, and local minimums and maximums. I then entered the eight data points into my graphing calculator by following the directions that were explained in class. I was able to get the equation: y=-0.004x^4-0.004x^3+0.106x^2+0.119x-0.472 by using my calculator. I then used the points that I had found before to write down my x-intercepts and y-intercepts. These were found by looking at the graph.

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In order to get the coordinate values of the function. I put the photo over a graph in google docs and made the photo transparent. I then plugged these coordinate values I got from the graph into the "stat", "edit:1" table on my calculator. Then I used "stat", "calc: 7" in order to get the coefficient values of the function's equation. I know the y-int is .2982126165 because that is the e value the calculator gave me. I determined that since the end behavior of the function was as x approaches infinity, f(x) approaches infinity, and as x approaches -infinity, f(x) approaches infinity; that the range had a minimum value and that the domain was all real numbers. I then used the minimum calculation on my calculator to determine that the minimum value was -1.975385.

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Equation: y=-0.004x^4-0.004x^3+0.106x^2+0.119x-0.472
x-intercepts: (-4.75, 0), (-3, 0), (1.75, 0)
y-intercepts: (0, -0.5)

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Equation: .003607825x^4 + -.181139178x^2 + .2982126165

Y-int= .2982126165

Domain: (-infinity, infinity)
Range: (-1.975385, infinity)

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This is a photo I took of a sunset in Vermont this summer! The curve created by the mountain tops represents a function of a higher degree.

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I began by looking at the photo and tracing it on the axis´ I made. I then found the zeros which were -10,-2,2,and 4. After that, I put in the zeros and then made a partial equation which was, f(x)=a(x+10)(x+2)(x-2)(x-4).
I then found the y intercept (0,-1) and plugged it into the equation so I could find a. My equation was then
-1=a(0+10)(0+2)(0-2)(0-4).
After that was solved, the answer I got was a=-1/160. My final equation was
f(x)=-1/160(x+10)(x+2)(x-2)(x-4).

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My graph with intercepts, function, and end behavior.

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I started by tracing the mountain picture. I then found three points on the graph that were easy to plug into a graphing calculator.(-5,0)(2.5,0)(0,-5) I then plugged these numbers into a graphing calculator the same way that was explained in class. The equation came out to x^4-4x^3+6x^2-4x-5. The x and y intercepts were already found by choosing the points earlier above.

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Equation= x^4-4x^3+6x^2-4x-5,
Y int: (-5,0)(2.5,0), X int: (0,-5)

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This is a sunset placed in between a mountain range. The mountains represent a fourth degree polynomial function.

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I began by hand drawing a graphing and then tracing the top of the sign on it.. Since the top of the sign was a fourth degree polynomial and the two curves were almost exactly identical this was easy to graph. I then used the points from my graph ((-5,0) (-2,0) (5,0) (2,0) (0,3)) to find an equation using the method taught in class via a graphing calculator. Since my coordinates were all real whole numbers my equation came out to be f(x) = .03x^4 - .87x^2 + 3 and did not need any extra conversions. I already knew my zeros and intercepts by looking at my hand drawn graph, but to ensure they are correct I checked them using the intercept option in the graphing calculator.

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Equation : f(x) = .03x^4 - .87x^2 + 3
Zeros : -5, -3, 2, 5
X-intercepts : (-5,0) (-2,0) (5,0) (2,0)
Y intercept : (0,3)
Graph: shown in image

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I began by placing a graph over the picture of my cat and outlining the 4th degree polynomial. From the polynomial I found my zeroes (x=0.3 and x= -3.3) and y-intercept (0,4.2). Then I pulled 5 points ((-3,3.5), (-2,2.8), (-1,2.6), (-0.5, 3.5), and (0.25,3.15)) and used the method done in class on Thursday. I plugged the points into my calculator and it gave me the values of the coefficients (in decimal form). From there, I converted the decimals into fractions and wrote my equation. Using the equation and graph I found my domain (-∞,∞) and range (-∞, 4.2), as well as my end behavior (as f(x) -> -∞, x ->-∞; as f(x) -> -∞, x ->∞) .

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This is the equation, graph, domain, range, and end behavior.

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I started by placing the snake picture on a graph and determining the zeros and y intercept. The zeros were x=-8 (multiplicity 3) and x=5 (multiplicity 1), and the y intercept was at (0,-10). Next, I wrote out f(x)=a(x+8)^3 (x-5) and found that a=1/256 by plugging the y intercept in. I wrote out the end behavior and then drew the function on the graph. The domain was (-∞,∞), and to find the range I used calc minimum to find the absolute minimum of the polynomial, which was (1.75,-11.77), so the range was [-11.77,∞).

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The graph, equation, intercepts, domain and range, and end behavior.

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I took the picture and traced it lightly onto a graph to determine where I could make the zeros, -5.5, -1.5, 1.5, and 5.5. Then, I took those zeros, turned them into factors, and multiplied them together to get a final equation. I plugged in some x values to get other points, and I had my graph.

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This is the equation and intercepts--as well as a few extra points--of my graph (with a roughly traced version of the picture for reference. I can't draw cats).

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After putting the chain over a graph, the end of the chain was at the point (0,-6) and the other end of the chain was at (0,6). The other zeros were at (0,2) and (0,-2). I drew a picture of the chain with those points. Using the zeroes, I created an equation however I was missing the leading coefficient. After plugging the y intercept (3,0) into the equation, I was able to figure out “a”, the leading coefficient.

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The equation and zeroes of my graph

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First I traced the rough outlines of the mountain into my graph. Then I got points from -10 to 10 and plug the numbers into my calculator and used the method showed on thursday to find the equation. I used the trace function of the graph to find the x and y intercepts.

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My graph and equations along with the x and y intercepts

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My graph and equations along with the x and y intercepts

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I used this part of my picture for the graph and equation.

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I used this part of my picture for the graph and equation.

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I drew a line over the water from the waterfall and drew a graph over the photo. I rotated the line to be symmetrical on the y axis. The y intercept was (0,3) which meant that there was a plus 3 shift at the end of the equation. I found the zero (7,0) by looking at the graph. I used the equation shown below to find the leading coefficient. The x intercept is (7,0) and the y intercept is (0,3) based on the graph.

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-(3/343)(x^3) + 3
y intercept is (0,3)
x intercept is (7,0)

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With the five points I put them into the calculator stat edit enter. Then stat calc for a quartic function.  After that I made the equation show up on Y1 with vars then y vars-function. The function goes on forever at the x-axis so its neg infinity to infinity. Then the function comes up barely over 0 at the y-axis then goes on forever downward which gave me the range of neg inf -> +0.18.

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This is the equation I got using the caculator and the domain and range.

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This is the equation I got using the caculator and the domain and range.

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This is a picture of a waterfall I took. The falling water represents the higher degree polynomial.

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This is a photo on the beach during a sunset that I took in Cape Cod over the summer. The clouds in the photo would represent a higher degree.

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This is my cat griffin; he is a very fluffy example of a 3rd degree polynomial, from his head to his feet.

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Jack haber f-block
This is a chain that I wear almost every day the positioning of the chain creates a higher degree polynomial function

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This is a photograph of an electric guitar. The left side specifically, demonstrates a 4th degree polynomial.

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This is my cat Tiny in a basket. The top of the basket looks like it could fit onto the graph of a fourth degree polynomial.

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This is a picture I took of a sign at a corn maze I went to! The top of the sign creates a perfect 4th degree polynomial!

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This is a picture of my cat. The shape of his ears make a 4th degree polynomial.

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This is a picture of a deceased eastern garter snake that I took in my front yard. The flexible backbone of this small snake is a great example of a natural polynomial function of at least 4th degree.

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This is a photo of an item from my everyday life, my glasses. The bottom and top frame of my glasses represent a higher degree polynomial function. For this project I will be using the bottom frame as the function is more distinct.

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I took this picture while skiing at Berkshire East, and in the background you can see multiple mountains. These mountains can demonstrate polynomials of a high degree.

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I chose this photo because it was taken in Nantucket which is my family’s favorite vacation spot. The dunes and hills are curved in a way that perfectly demonstrates a high degree function.

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Mountains of the Himalayas-Sujal Manandhar-F Block

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Mountains of the Himalayas-Sujal Manandhar-F Block

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I had this photo from 2018 where I took a picture of the Himalayas from a plane window. I though the peaks of the mountains showed a polynomial of a higher degree.

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Reflection of Trees on MHC Lake- Alex Martin, (F Block)

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Reflection of Trees on MHC Lake- Alex Martin, (F Block)

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I took this picture about one week ago as I was walking my dog around the Mount Holyoke Lake. It is a clear photo that shows a polynomial of a degree of three or more.

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Mountains at Lake WIlloughby - Ruth Pietrzykowski, F-Block

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Mountains at Lake WIlloughby - Ruth Pietrzykowski, F-Block

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I took this photo in August 2017 at Lake Willoughby in Westmore, Vermont. I found it in my camera roll and thought it would be a perfect example of a polynomial for this project!